23.7 - Solved Examples on Integration Using Trigonometric Identities

In the previous section, we saw the application of trigonometric identities in the process of integration. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 23.9
Find the following integrals:
(i) $\small{\int{\left[\sin 4x \sin 8x \right]dx}}$

(ii) $\small{\int{\left[\sin x \sin 2x \sin 3x \right]dx}}$

(iii) $\small{\int{\left[\cos 2x \cos 4x \cos 6x \right]dx}}$

Solution:
Part (i):
1. We have the identity:
$\small{\cos A\,-\,\cos B\,=\,-2 \sin \left(\frac{A+B}{2} \right) \sin \left(\frac{A - B}{2} \right)}$
• From this, we get: $\small{\sin \left(\frac{A+B}{2} \right) \sin \left(\frac{A - B}{2} \right) \,=\,\frac{-(\cos A\,-\,\cos B)}{2}}$

2. So for our present problem, we can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{A+B}{2}}    & {~=~}    &{8x}    \\
{~\color{magenta}    2    }    &{{}}    &{{\frac{A-B}{2}}}    & {~=~}    &{4x}    \\
\end{array}}$

• Solving the two equations, we get:
A = 12x and B = 4x

• So the given function can be rearranged as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin 4x \sin 8x\right] \, dx}}    & {~=~}    &{\int{\left[\frac{-(\cos 12x\,-\,\cos 4x)}{2}\right] \, dx}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{\cos 4x \,-\,\cos 12x}{2}\right] \, dx}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \int{\left[\cos 4x \right] \, dx}~-~\frac{1}{2} \int{\left[\cos 12x \right] \, dx}}    \\
\end{array}}$

3. The R.H.S has two terms. We will consider each term separately.
First term:
For this term, we use the method of substitution.
(i) The derivative of (4x) is 4.
• So we put u = 4x
⇒ $\small{\frac{du}{dx}~=~4}$
⇒ 4 dx = du

(ii) So we want:
$\small{\frac{1}{2} \int{\left[\frac{4 \cos 4x}{4} \right]dx}~=~\frac{1}{2} \int{\left[\frac{\cos u}{4} \right]du}}$

• This integration gives:
$\small{\frac{1}{2} \frac{\sin u}{4}\,+\,C_1~=~ \frac{\sin 4x}{8}\,+\,C_1}$

Second term:
For this term also, we use the method of substitution.
(i) The derivative of (12x) is 12.
• So we put u = 12x
⇒ $\small{\frac{du}{dx}~=~12}$
⇒ 12 dx = du

(ii) So we want:
$\small{\frac{1}{2} \int{\left[\frac{12 \cos 12x}{12} \right]dx}~=~\frac{1}{2} \int{\left[\frac{\cos u}{12} \right]du}}$

• This integration gives:
$\small{\frac{1}{2} \frac{\sin u}{12}\,+\,C_2~=~ \frac{\sin 12x}{24}\,+\,C_2}$

4. Now, based on step 2, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin 4x \sin 8x \right]dx}}    & {~=~}    &{\frac{\sin 4x}{8}\,+\,C_1~-~\frac{\sin 12x}{24}\,-\,C_2}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{\sin 4x}{8}\,-\,\frac{\sin 12x}{24}\,+\,C}    \\
\end{array}}$                           

• Note that, the constants C₁, C₂ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

Part (ii): $\small{\int{\left[\sin x \sin 2x \sin 3x \right]dx}}$
• The given function can be rearranged into another form. We will do the rearrangement in three stages.

Stage I:
• x is odd, 2x is even and 3x is odd.
• We must do only two operations:
   ♦ add odd to odd
   ♦ add even to even
• Only then we will be able to divide by 2
• So we will choose (sin x sin 3x) for stage I

1. We have the identity:
$\small{\cos A\,-\,\cos B\,=\,-2 \sin \left(\frac{A+B}{2} \right) \sin \left(\frac{A - B}{2} \right)}$
• From this, we get: $\small{\sin \left(\frac{A+B}{2} \right) \sin \left(\frac{A - B}{2} \right) \,=\,\frac{-(\cos A\,-\,\cos B)}{2}}$

2. So for our present problem, we can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{A+B}{2}}    & {~=~}    &{3x}    \\
{~\color{magenta}    2    }    &{{}}    &{{\frac{A-B}{2}}}    & {~=~}    &{1x}    \\
\end{array}}$

• Solving the two equations, we get:
A = 4x and B = 2x

• So (sin x sin 3x) can be rearranged as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\sin x \sin 3x}    & {~=~}    &{\frac{-(\cos 4x\,-\,\cos 2x)}{2}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{\cos 2x \,-\,\cos 4x}{2}}    \\
\end{array}}$

• Now the given function can be rearranged as:
sin x sin 2x sin 3x = (sin x sin 3x) sin 2x
= $\small{\left(\frac{\cos 2x\,-\,\cos 4x}{2} \right)\sin 2x}$
= $\small{\frac{\sin 2x \cos 2x\,-\,\sin 2x\cos 4x}{2} }$
• In stage II, we will rearrange the first term (sin 2x cos 2x)

Stage II: sin 2x cos 2x

1. We have the identity:
$\small{\sin 2A\,=\,2 \sin A \cos A}$
• From this, we get: $\small{\sin A \cos A\,=\,\frac{\sin 2A}{2}}$

2. So for our present problem, we can write: A = 2x
• So (sin 2x cos 2x) can be rearranged as: $\small{\frac{\sin 4x}{2}}$

• Now the given function can be rearranged as:
sin x sin 2x sin 3x = (sin x sin 3x) sin 2x
= $\small{\left(\frac{\cos 2x\,-\,\cos 4x}{2} \right)\sin 2x}$
= $\small{\frac{\sin 2x \cos 2x\,-\,\sin 2x\cos 4x}{2} }$
= $\small{\frac{\frac{\sin 4x}{2}\,-\,\sin 2x\cos 4x}{2} }$
• In stage III, we will rearrange (sin 2x cos 4x)

Stage III: (sin 2x cos 4x)

1. We have the identity:
$\small{\sin A\,-\,\sin B\,=\,-2 \cos \left(\frac{A+B}{2} \right) \sin \left(\frac{A - B}{2} \right)}$
• From this, we get: $\small{\cos \left(\frac{A+B}{2} \right) \sin \left(\frac{A - B}{2} \right) \,=\,\frac{\sin A\,-\,\sin B)}{2}}$

2. So for our present problem, we can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{A+B}{2}}    & {~=~}    &{4x}    \\
{~\color{magenta}    2    }    &{{}}    &{{\frac{A-B}{2}}}    & {~=~}    &{2x}    \\
\end{array}}$

• Solving the two equations, we get:
A = 6x and B = 2x

• So (sin 2x cos 4x) can be rearranged as:
$\small{\frac{\sin 6x \,-\, \sin 2x}{2}}$

• Now the given function can be rearranged in the final form:
sin x sin 2x sin 3x = (sin x sin 3x) sin 2x
= $\small{\left(\frac{\cos 2x\,-\,\cos 4x}{2} \right)\sin 2x}$
= $\small{\frac{\sin 2x \cos 2x\,-\,\sin 2x\cos 4x}{2} }$
= $\small{\frac{\frac{\sin 4x}{2}\,-\,\sin 2x\cos 4x}{2} }$
= $\small{\frac{\frac{\sin 4x}{2}\,-\,\left(\frac{\sin 6x \,-\, \sin 2x}{2} \right)}{2} }$
= $\small{\frac{\sin 4x}{4}\,-\,\frac{\sin 6x}{4}\,+\,\frac{\sin 2x}{4} }$

• Now we can begin the integration process.
• Based on the solved examples that we discussed so far, we can write two formulas:

(i) $\small{\int{\left[\sin mx \right]dx}~=~\frac{-\cos mx}{m}}$

(ii) $\small{\int{\left[\cos mx \right]dx}~=~\frac{\sin mx}{m}}$

• The above two formulas can be used directly while solving problems. So we can easily write the integral:
$\small{\int{\left[\frac{\sin 4x}{4}\,-\,\frac{\sin 6x}{4}\,+\,\frac{\sin 2x}{4} \right]dx}}$

$\small{~=~\left[\frac{-\cos 4x}{4(4)}\,+\,C_1 \right]\,-\,\left[\frac{-\cos 6x}{4(6)}\,+\,C_2 \right]\,+\,\left[\frac{-\cos 2x}{4(2)}\,+\,C_3 \right]}$

$\small{~=~\frac{1}{4}\left[\frac{-\cos 4x}{4}\,+\,\frac{\cos 6x}{6}\,-\,\frac{\cos 2x}{2}\right]\,+\,C}$

• Note that, the constants C₁, C₂ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

Part (iii): $\small{\int{\left[\cos 2x \cos 4x \cos 6x \right]dx}}$
• The given function can be rearranged into another form. We will do the rearrangement in three stages.

Stage I:
• 2x is even, 4x is even and 6x is even.
• We must do only two operations:
   ♦ add odd to odd
   ♦ add even to even
• Only then we will be able to divide by 2
• Here all terms are even. So we can choose in any order we like. We will choose (cos 2x cos 4x) for stage I

1. We have the identity:
$\small{\cos A\,+\,\cos B\,=\,2 \cos \left(\frac{A+B}{2} \right) \cos \left(\frac{A - B}{2} \right)}$
• From this, we get: $\small{\cos \left(\frac{A+B}{2} \right) \cos \left(\frac{A - B}{2} \right) \,=\,\frac{\cos A\,+\,\cos B}{2}}$

2. So for our present problem, we can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{A+B}{2}}    & {~=~}    &{4x}    \\
{~\color{magenta}    2    }    &{{}}    &{{\frac{A-B}{2}}}    & {~=~}    &{2x}    \\
\end{array}}$

• Solving the two equations, we get:
A = 6x and B = 2x

• So (cos 2x cos 4x) can be rearranged as:
$\small{\cos 2x \cos 4x~=~\frac{\cos 6x \,+\,\cos 2x}{2}}$

• Now the given function can be rearranged as:
cos 2x cos 4x cos 6x = (cos 2x cos 4x) cos 6x
= $\small{\left(\frac{\cos 6x\,+\,\cos 2x}{2} \right)\cos 6x}$
= $\small{\frac{\cos^2 6x\,+\,\cos 2x\cos 6x}{2} }$
• In stage II, we will rearrange the first term (cos²6x)

Stage II: cos²6x

1. We have the identity: $\small{\cos 2A \,=\, 2 \cos^2 A \,-\, 1}$
• From this, we get: $\small{\cos^2 A \,=\,\frac{1\,+\,\cos 2A}{2}}$

2. So for our present problem, we can write:
$\small{\cos^2 6 x \,=\,\frac{1\,+\,\cos 12x}{2}}$

• Now the given function can be rearranged as:
cos 2x cos 4x cos 6x = (cos 2x cos 4x) cos 6x
= $\small{\left(\frac{\cos 6x\,+\,\cos 2x}{2} \right)\cos 6x}$
= $\small{\frac{\cos^2 6x\,+\,\cos 2x\cos 6x}{2} }$
= $\small{\frac{\frac{1\,+\,\cos 12x}{2}\,+\,\cos 2x\cos 6x}{2} }$
• In stage III, we will rearrange (cos 2x cos 6x)

Stage III: (cos 2x cos 6x)

1. We have the identity:
$\small{\cos A\,+\,\cos B\,=\,2 \cos \left(\frac{A+B}{2} \right) \cos \left(\frac{A - B}{2} \right)}$
• From this, we get: $\small{\cos \left(\frac{A+B}{2} \right) \cos \left(\frac{A - B}{2} \right) \,=\,\frac{\cos A\,+\,\cos B}{2}}$

2. So for our present problem, we can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{A+B}{2}}    & {~=~}    &{6x}    \\
{~\color{magenta}    2    }    &{{}}    &{{\frac{A-B}{2}}}    & {~=~}    &{2x}    \\
\end{array}}$

• Solving the two equations, we get:
A = 8x and B = 4x

• So (cos 2x cos 6x) can be rearranged as:
$\small{\cos 2x \cos 6x~=~\frac{\cos 8x \,+\,\cos 4x}{2}}$

• Now the given function can be rearranged in the final form:
cos 2x cos 4x cos 6x = (cos 2x cos 4x) cos 6x
= $\small{\left(\frac{\cos 6x\,+\,\cos 2x}{2} \right)\cos 6x}$
= $\small{\frac{\cos^2 6x\,+\,\cos 2x\cos 6x}{2} }$
= $\small{\frac{\frac{1\,+\,\cos 12x}{2}\,+\,\cos 2x\cos 6x}{2} }$
= $\small{\frac{\frac{1\,+\,\cos 12x}{2}\,+\,\frac{\cos 8x \,+\,\cos 4x}{2}}{2} }$
= $\small{\frac{\frac{1\,+\,\cos 12x\,+\,\cos 8x\,+\,\cos 4x}{2}}{2} }$
= $\small{\frac{1}{4}\,+\,\frac{\cos 12x}{4}\,+\,\frac{\cos 8x}{4}\,+\,\frac{\cos 4x}{4}}$

• Now we can begin the integration process.
• Based on the solved examples that we discussed so far, we can write two formulas:

(i) $\small{\int{\left[\sin mx \right]dx}~=~\frac{-\cos mx}{m}}$

(ii) $\small{\int{\left[\cos mx \right]dx}~=~\frac{\sin mx}{m}}$

• The above two formulas can be used directly while solving problems. So we can easily write the integral:
$\small{\int{\left[\frac{1}{4}\,+\,\frac{\cos 12x}{4}\,+\,\frac{\cos 8x}{4}\,+\,\frac{\cos 4x}{4} \right]dx}}$

$\small{~=~\left[\frac{x}{4}\,+\,C_1 \right]\,+\,\left[\frac{\sin 12x}{4(12)}\,+\,C_2 \right]\,+\,\left[\frac{\sin 8x}{4(8)}\,+\,C_3 \right]\,+\,\left[\frac{\sin 4x}{4(4)}\,+\,C_4 \right]}$

$\small{~=~\frac{1}{4}\left[x\,+\,\frac{\sin 12x}{12}\,+\,\frac{\sin 8x}{8}\,+\,\frac{\sin 4x}{4}\right]\,+\,C}$

• Note that, the constants C₁, C₂ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

---

The link below gives a few more miscellaneous examples:

Exercise 23.3

---

In the next section, we will see a few more solved examples.

Previous

Contents

Next

Copyright©2025 Higher secondary mathematics.blogspot.com

23.6 - Integration Using Trigonometric Identities

In the previous section, we saw some solved examples demonstrating integration by substitution. In this section, we will see the application of trigonometric identities in the process of integration. We have already seen some examples in the previous section. For example, in the solved example 23.7(ii) of the previous section, we used the identity:
$\small{\sin A \,-\, \sin B ~=~2 \cos \left(\frac{A+B}{2} \right) \sin \left(\frac{A-B}{2} \right)}$

Now we will see some advanced problems

Solved example 23.8
Find the following integrals:
(i) $\small{\int{\left[\cos^2 x \right]dx}}$

(ii) $\small{\int{\left[\sin 2x \cos 3x \right]dx}}$

(iii) $\small{\int{\left[\sin^3 x \right]dx}}$

(iv) $\small{\int{\left[\sin 3x \cos 4x \right]dx}}$

Solution:
Part (i):
1. We have the identity: $\small{\cos 2A \,=\, 2 \cos^2 A \,-\, 1}$
• From this, we get: $\small{\cos^2 A \,=\,\frac{1\,+\,\cos 2A}{2}}$

2. So for our present problem, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\cos^2 x \right]dx}}    & {~=~}    &{\int{\left[\frac{1\,+\,\cos 2x}{2}\right] \, dx}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \int{\left[1 \right] \, dx}~+~\frac{1}{2} \int{\left[\cos 2x \right] \, dx}}    \\
\end{array}}$

3. The R.H.S has two terms. We will consider each term separately.
First term:
• This term is easy. We can directly write:
$\small{\frac{1}{2} \int{\left[1 \right] \, dx~=~\frac{x}{2}\,+\,C_1}}$

Second term:
For this term, we use the method of substitution.
(i) The derivative of (2x) is 2.
• So we put u = 2x
⇒ $\small{\frac{du}{dx}~=~2}$
⇒ 2 dx = du

(ii) So we want:
$\small{\frac{1}{2} \int{\left[\frac{2 \cos 2x}{2} \right]dx}~=~\frac{1}{2} \int{\left[\frac{\cos u}{2} \right]du}}$

• This integration gives:
$\small{\frac{1}{2} \frac{\sin u}{2}\,+\,C_2~=~ \frac{\sin 2x}{4}\,+\,C_2}$

4. Now, based on step 2, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\cos^2 x\right]dx}}    & {~=~}    &{\frac{x}{2}\,+\,C_1~+~\frac{\sin 2x}{4}\,+\,C_2}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2}\,+\,\frac{\sin 2x}{4}\,+\,C}    \\
\end{array}}$                           

• Note that, the constants C₁, C₂ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

Part (ii):
1. We have the identity:
$\small{\sin A\,-\,\sin B\,=\,2 \cos\left(\frac{A+B}{2} \right) \sin \left(\frac{A - B}{2} \right)}$
• From this, we get: $\small{\cos\left(\frac{A+B}{2} \right) \sin \left(\frac{A - B}{2} \right) \,=\,\frac{\sin A\,-\,\sin B}{2}}$

2. So for our present problem, we can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{A+B}{2}}    & {~=~}    &{3x}    \\
{~\color{magenta}    2    }    &{{}}    &{{\frac{A-B}{2}}}    & {~=~}    &{2x}    \\
\end{array}}$

• Solving the two equations, we get:
A = 5x and B = 1x

• So the given function can be rearranged as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin 2x \cos 3x\right] \, dx}}    & {~=~}    &{\int{\left[\frac{\sin 5x\,-\,\sin x}{2}\right] \, dx}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \int{\left[\sin 5x \right] \, dx}~-~\frac{1}{2} \int{\left[\sin x \right] \, dx}}    \\
\end{array}}$

3. The R.H.S has two terms. We will consider each term separately.
First term:
For this term, we use the method of substitution.
(i) The derivative of (5x) is 5.
• So we put u = 5x
⇒ $\small{\frac{du}{dx}~=~5}$
⇒ 5 dx = du

(ii) So we want:
$\small{\frac{1}{2} \int{\left[\frac{5 \sin 5x}{5} \right]dx}~=~\frac{1}{2} \int{\left[\frac{\sin u}{5} \right]du}}$

• This integration gives:
$\small{\frac{1}{2} \frac{-\cos u}{5}\,+\,C_1~=~ \frac{-\cos 5x}{10}\,+\,C_1}$

Second term:
• This term is easy. We can directly write:
$\small{\frac{1}{2} \int{\left[\sin x \right] \, dx~=~\frac{-\cos x}{2}\,+\,C_2}}$

4. Now, based on step 2, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin 2x \cos 3x \right]dx}}    & {~=~}    &{\frac{-\cos 5x}{10}\,+\,C_1~-~\frac{-\cos x}{2}\,+\,C_2}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{-\cos 5x}{10}\,+\,\frac{\cos x}{2}\,+\,C}    \\
\end{array}}$                           

• Note that, the constants C₁, C₂ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

Part (iii):
1. The given expression can be rearranged as follows:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\sin^3 x}    & {~=~}    &{\sin x \,\sin^2 x}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\sin x(1\,-\,\cos^2 x)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\sin x \,-\,\sin x \, \cos^2 x}    \\
\end{array}}$

2. So for our present problem, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[ \sin^3 x \right]dx}}    & {~=~}    &{\int{\left[\sin x \,-\,\sin x \, \cos^2 x \right]dx}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\sin x \right]}dx~-~\int{\left[\sin x \, \cos^2 x \right]}dx}    \\
\end{array}}$

3. The R.H.S has two terms. We will consider each term separately.
First term:
• This term is easy. We can directly write:
$\small{\int{\left[\sin x \right] }\, dx~=~-\cos x \,+\,C_1}$

Second term:
For this term, we use the method of substitution.
(i) The derivative of (cos x) is −sin x.
• So we put u = cos x
⇒ $\small{\frac{du}{dx}~=~-\sin x}$
⇒ (−sin x)dx = du

(ii) So we want:
$\small{\int{\left[(-1)(-1) \sin x \, \cos^2 x \right]dx}~=~ \int{\left[(-1) u^2 \right]du}}$

• This integration gives:
$\small{(-1) \frac{u^3}{3}\,+\,C_2~=~ (-1)\frac{\cos^3 x}{3}\,+\,C_2}$

4. Now, based on step 2, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin^3 x\right]dx}}    & {~=~}    &{-\cos x \,+\,C_1~-~ (-1)\frac{\cos^3 x}{3}\,+\,C_2}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{-\cos x ~+~ \frac{\cos^3 x}{3}\,+\,\,C}    \\
\end{array}}$

• Note that, the constants C₁, C₂ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

Alternate method:

1. We have the identity: $\small{\sin 3A\,=\,3 \sin A \,-\,4 \sin^3 A}$
• From this, we get: $\small{\sin^3 A \,=\,\frac{3 \sin A \,-\,\sin 3A}{4}}$

2. So for our present problem, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin^3 x \right]dx}}    & {~=~}    &{\int{\left[\frac{3 \sin x \,-\,\sin 3x}{4}\right] \, dx}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{3}{4} \int{\left[\sin x \right] \, dx}~-~\frac{1}{4} \int{\left[\sin 3x \right] \, dx}}    \\
\end{array}}$

3. The R.H.S has two terms. We will consider each term separately.
First term:
• This term is easy. We can directly write:
$\small{\frac{3}{4} \int{\left[\sin x \right] \, dx~=~\frac{-3 \cos x}{4}\,+\,C_1}}$

Second term:
For this term, we use the method of substitution.
(i) The derivative of (3x) is 3.
• So we put u = 3x
⇒ $\small{\frac{du}{dx}~=~3}$
⇒ 3 dx = du

(ii) So we want:
$\small{\frac{1}{4} \int{\left[\frac{3 \sin 3x}{3} \right]dx}~=~\frac{1}{4} \int{\left[\frac{\sin u}{3} \right]du}}$

• This integration gives:
$\small{\frac{1}{4} \frac{-\cos u}{3}\,+\,C_2~=~ \frac{-\cos 3x}{12}\,+\,C_2}$

4. Now, based on step 2, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin^3 x\right]dx}}    & {~=~}    &{\frac{-3 \cos x}{4}\,+\,C_1~-~\frac{-\cos 3x}{12}\,+\,C_2}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{-3 \cos x}{4}\,+\,\frac{\cos 3x}{12}\,+\,C}    \\
\end{array}}$

• Note that, the constants C₁, C₂ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

---

We see that, the results obtained by the two methods are different. But their equality can be proved by using trigonometric identities. This is shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{-3 \cos x}{4}\,+\,\frac{\cos 3x}{12}}    & {~=~}    &{\frac{-3 \cos x}{4}\,+\,\frac{4 \cos^3 x\,-\,3 \cos x}{12}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{-3 \cos x}{4}\,+\,\frac{4 \cos^3 x}{12}\,-\,\frac{3 \cos x}{12}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{-3 \cos x}{4}\,+\,\frac{\cos^3 x}{3}\,-\,\frac{\cos x}{4}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{-4 \cos x}{4}\,+\,\frac{\cos^3 x}{3}}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{-\cos x\,+\,\frac{\cos^3 x}{3}}    \\
\end{array}}$

---

Part (iv):
1. We have the identity:
$\small{\sin A \,-\,\sin B\,=\,2 \cos \left(\frac{A+B}{2} \right)\,\sin \left(\frac{A-B}{2} \right)}$
• From this, we get: $\small{\cos\left(\frac{A+B}{2} \right) \sin \left(\frac{A - B}{2} \right) \,=\,\frac{\sin A\,-\,\sin B}{2}}$

2. So for our present problem, we can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{A+B}{2}}    & {~=~}    &{4x}    \\
{~\color{magenta}    2    }    &{{}}    &{{\frac{A-B}{2}}}    & {~=~}    &{3x}    \\
\end{array}}$

• Solving the two equations, we get:
A = 7x and B = 1x

• So the given function can be rearranged as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin 3x \cos 4x\right] \, dx}}    & {~=~}    &{\int{\left[\frac{\sin 7x\,-\,\sin x}{2}\right] \, dx}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \int{\left[\sin 7x \right] \, dx}~-~\frac{1}{2} \int{\left[\sin x \right] \, dx}}    \\
\end{array}}$

3. The R.H.S has two terms. We will consider each term separately.
First term:
For this term, we use the method of substitution.
(i) The derivative of (7x) is 7.
• So we put u = 7x
⇒ $\small{\frac{du}{dx}~=~7}$
⇒ 7 dx = du

(ii) So we want:
$\small{\frac{1}{2} \int{\left[\frac{7 \sin 7x}{7} \right]dx}~=~\frac{1}{2} \int{\left[\frac{\sin u}{7} \right]du}}$

• This integration gives:
$\small{\frac{1}{2} \frac{-\cos u}{5}\,+\,C_1~=~ \frac{-\cos 7x}{14}\,+\,C_1}$

Second term:
• This term is easy. We can directly write:
$\small{\frac{1}{2} \int{\left[\sin x \right] \, dx~=~\frac{-\cos x}{2}\,+\,C_2}}$

4. Now, based on step 2, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin 3x \cos 4x \right]dx}}    & {~=~}    &{\frac{-\cos 7x}{14}\,+\,C_1~-~\frac{-\cos x}{2}\,+\,C_2}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{-\cos 7x}{14}\,+\,\frac{\cos x}{2}\,+\,C}    \\
\end{array}}$                           

• Note that, the constants C₁, C₂ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

---

In the next section, we will see a few more solved examples.

Previous

Contents

Next

Copyright©2025 Higher secondary mathematics.blogspot.com

18.7 - Properties of Inverse Trigonometric Functions

In the previous section, we saw some solved examples on the six inverse trigonometric functions. In this section, we will see the properties of inverse trigonometric functions.

First we will see how the inputs and outputs are to be denoted. It can be written in 3 steps:  

1. In the discussions so far in this chapter, we gave primary importance to the original trigonometric function.
• For example, the original trigonometric function was:
y = f(x) = sin x
    ♦ Input was denoted as ‘x’.
    ♦ Output was denoted as ‘y’.
2. From the original trigonometric function, we derived the inverse trigonometric function.
• For example:
x = f(y) = sin^-1 y
    ♦ Input was denoted as ‘y’.
    ♦ Output was denoted as ‘x’.
3. We have learned the basic details about inverse trigonometric functions. We can now treat them independently. So from now on wards, the inverse trigonometric function will be our primary function.
    ♦ Input for the inverse trigonometric function will be denoted as ‘x’.
    ♦ Output for the inverse trigonometric function will be denoted as ‘y’.

---

Now we will see an interesting case. It can be written in 3 steps:
1. In the previous chapter, we saw composite functions. Let us recall the details:
(i) f: A→B and g: B→C are two functions.
• Then the composite function g(f(x)) connects the domain of f with the codomain of g.
• So we can write:
    ♦ Domain of g(f(x)) is A
    ♦ Codomain of g(f(x)) is C
(ii) Suppose that, g is the inverse of f.
• Then f is mapped from A to B and g is mapped from B to A.
• In such a situation, we can write:
    ♦ Domain of g(f(x)) is A
    ♦ Codomain of g(f(x)) is also A
This is because, g(f(x)) connects the domain of f with the codomain of g.
• Also, g(f(x)) is an identity function. Whatever value we give as input, the output will be that same value.
(Recall that, function of “the inverse of that function” is an identity function)
• Some examples can be seen here.
(iii) Similarly, we can write about f(g(x)):
    ♦ Domain of f(g(x)) is B .
    ♦ Codomain of f(g(x)) is also B.
• f(g(x)) is also an identity function. Whatever value we give as input, the output will be that same value.
2. Now we will apply the above information on sin function and it’s inverse.
(i) Let us write the domain and codomain:
    ♦ sin function is mapped from $\left[-{\frac{\pi}{2}, \frac{\pi}{2}} \right]$ to [-1,1]
    ♦ sin^-1 function is mapped from [-1,1] to $\left[-{\frac{\pi}{2}, \frac{\pi}{2}} \right]$
(ii) sin(sin^-1 x) will connect the domain of sin^-1 with the codomain of sin.
• That means, sin(sin^-1 x) is mapped from [-1,1] to [-1,1]
• In other words, sin(sin^-1 x) is a function on [-1,1]
• Also, sin(sin^-1 x) is an identity function. Whatever value we give as input, the output will be that same value.
(iii) sin^-1(sin x) will connect the domain of sin with the codomain of sin^-1.
• That means, sin^-1(sin x) is mapped from $\left[-{\frac{\pi}{2}, \frac{\pi}{2}} \right]$ to $\left[-{\frac{\pi}{2}, \frac{\pi}{2}} \right]$
• In other words, sin^-1(sin x) is a function on $\left[-{\frac{\pi}{2}, \frac{\pi}{2}} \right]$.
• Also, sin^-1(sin x) is an identity function. Whatever value we give as input, the output will be that same value.
3. The above step 2 is applicable to all six trigonometric functions. So we can write 12 identity functions:

$\begin{array}{cc}{}    &{\text{(i)}}    &{\sin \left(\sin^{-1} (x) \right)}    &{\text{(vii)}}    &{\sin^{-1} \left(\sin(x) \right)}    &{}\\
{}    &{\text{(ii)}}    &{\cos \left(\cos^{-1} (x) \right)}    &{\text{(viii)}}    &{\cos^{-1} \left(\cos (x) \right)}    &{}\\
{}    &{\text{(iii)}}    &{\tan \left(\tan^{-1} (x) \right)}    &{\text{(ix)}}    &{\tan^{-1} \left(\tan (x) \right)}    &{}\\
{}    &{\text{(iv)}}    &{\csc \left(\csc^{-1} (x) \right)}    &{\text{(x)}}    &{\csc^{-1} \left(\csc (x) \right)}    &{}\\
{}    &{\text{(v)}}    &{\sec \left(\sec^{-1} (x) \right)}    &{\text{(xi)}}    &{\sec^{-1} \left(\sec (x) \right)}    &{}\\
{}    &{\text{(vi)}}    &{\cot \left(\cot^{-1} (x) \right)}    &{\text{(xii)}}    &{\cot^{-1} \left(\cot (x) \right)}    &{}\\
\end{array}                   
$

In all the above 12 cases, whatever value we give as input, the output will be that same value.

---

Now we will see 6 properties of inverse trigonometric functions.

Property I
• This has 3 parts:
\begin{array}{cc}{}    &{\text{(i)}}    &{\sin^{-1}\left(\frac{1}{x} \right)}    {}={}    &{\csc^{-1}x}    &{x \ge 1~\text{or}~x \le -1}\\
{}    &{\text{(ii)}}    &{\cos^{-1}\left(\frac{1}{x} \right)}    {}={}    &{\sec^{-1}x}    &{x \ge 1~\text{or}~x \le -1}\\
{}    &{\text{(iii)}}    &{\tan^{-1}\left(\frac{1}{x} \right)}    {}={}    &{\cot^{-1}x}    &{x > 0}\\
\end{array}

• Let us prove part (i):
$\begin{array}{ll}{}    &{\text{Let}~\sin^{-1} \left(\frac{1}{x} \right)}    & {~=~}    &{y~\color{magenta}{\text{- - - (A)}}}    &{} \\
{\implies}    &{\sin y}    & {~=~}    &{\frac{1}{x}}    &{} \\
{\implies}    &{\frac{1}{\csc y}}    & {~=~}    &{\frac{1}{x}}    &{} \\
{\implies}    &{\csc y}    & {~=~}    &{x}    &{} \\
{\implies}    &{\csc^{-1} x}    & {~=~}    &{y~\color{magenta}{\text{- - - (B)}}}    &{} \\
{\implies}    &{\csc^{-1} x}    & {~=~}    &{\sin^{-1} \left(\frac{1}{x} \right)~\color{magenta}{\text{- - - (C)}}}    &{} \\
\end{array}               
$

◼ Remarks:
• In line (A), we assume that $\sin^{-1}x$ = y
• In line B, we substitute for y, using the assumption in (A).
• When the substitution is done, we get line C. 

• Let us prove part (ii):
$\begin{array}{ll}{}    &{\text{Let}~\cos^{-1} \left(\frac{1}{x} \right)}    & {~=~}    &{y~\color{magenta}{\text{- - - (A)}}}    &{} \\
{\implies}    &{\cos y}    & {~=~}    &{\frac{1}{x}}    &{} \\
{\implies}    &{\frac{1}{\sec y}}    & {~=~}    &{\frac{1}{x}}    &{} \\
{\implies}    &{\sec y}    & {~=~}    &{x}    &{} \\
{\implies}    &{\sec^{-1} x}    & {~=~}    &{y~\color{magenta}{\text{- - - (B)}}}    &{} \\
{\implies}    &{\sec^{-1} x}    & {~=~}    &{\cos^{-1} \left(\frac{1}{x} \right)~\color{magenta}{\text{- - - (C)}}}    &{} \\
\end{array}               
$

◼ Remarks:
• In line (A), we assume that $\cos^{-1}x$ = y
• In line B, we substitute for y, using the assumption in (A).
• When the substitution is done, we get line C.

---

◼ In the same way, we can prove part (iii) also.
◼ Note:
For each of the three parts in property I, we see acceptable values of x. For example, for part (i), x ≥ 1 or x ≤ -1. We will see the details about such “acceptable values” in higher classes. At present, all we need to know is that, whenever we see $\csc^{-1} (x)$, we can put $\sin^{-1} \left(\frac{1}{x} \right)$ in it’s place.

---

In the next section, we will see the second and third properties of inverse trigonometric functions.

Previous

Contents

Next

Copyright©2024 Higher secondary mathematics.blogspot.com

Chapter 13.18 - Miscellaneous Examples

In the previous section, we completed a discussion on the derivatives of trigonometric functions. In this section, we will see some miscellaneous examples.

Solved example 13.19
Find the derivative of f from the first principle, where f is given by
(i) $f(x) = \frac{2x+3}{x-2} $    (ii) $f(x) = x + \frac{1}{x}$
Solution:
Part (i):
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{2(x+h) + 3}{(x+h) - 2}~–~ \frac{2x+3}{x-2}}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{2x + 2h + 3}{x+h - 2}~–~ \frac{2x+3}{x-2}}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{(2x + 2h + 3)(x-2) ~-~(x+h-2)(2x+3)}{(x+h - 2)(x-2)}}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{2x^2 - 4x + 2hx - 4h +3x - 6 - 2x^2 - 2hx + 4x - 3x - 3h +6}{(x+h - 2)(x-2)}}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{  - 4h - 3h }{(x+h - 2)(x-2)}}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{-7h}{h(x+h - 2)(x-2)} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{-7}{(x+h - 2)(x-2)} \right]}} &{} \\

{}&{}
& {~=~}& {\frac{-7}{(x - 2)(x-2)}} &{} \\

{}&{}
& {~=~}& {\frac{-7}{(x - 2)^2}} &{} \\

\end{array}$

Part (ii):
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[(x+h) + \frac{1}{(x+h)} \right]~-~\left[x + \frac{1}{x} \right]}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[\frac{(x+h)^2 + 1}{(x+h)} \right]~-~\left[\frac{x^2 + 1}{x} \right]}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{\left[(x+h)^2 + 1 \right]x~-~\left[(x^2 + 1)(x+h) \right]}{(x+h)x}}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[(x+h)^2 + 1 \right]x~-~\left[(x^2 + 1)(x+h) \right]}{(x+h)xh} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[x(x+h)^2 + x \right]~-~\left[x^3 + x^2 h + x + h \right]}{(x+h)xh} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[x^3 + 2 x^2 h + h^2 x + x \right]~-~\left[x^3 + x^2 h + x + h \right]}{(x+h)xh} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x^3 + 2 x^2 h + h^2 x + x ~-~x^3 - x^2 h - x - h }{(x+h)xh} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x^2 h + h^2 x - h }{(x+h)xh} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{h(x^2  + h x - 1) }{(x+h)xh} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x^2  + h x - 1 }{(x+h)x} \right]}} &{} \\

{}&{}
& {~=~}& {\frac{x^2  + (0) x - 1 }{(x+0)x}} &{} \\

{}&{}
& {~=~}& {\frac{x^2 - 1 }{x^2}} &{} \\

{}&{}
& {~=~}& {1 - \frac{1 }{x^2}} &{} \\

\end{array}$

Solved example 13.20
Find the derivative of f(x) from the first principle, where f(x) is
(i) sin x + cos x     (ii) x sin x
Solution:
Part (i):
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[\sin (x+h) + \cos (x+h) \right]~-~\left[\sin x + \cos x \right]}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin (x+h) + \cos (x+h) - \sin x - \cos x}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin (x+h) + \cos (x+h) - \sin x - \cos x}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[\sin (x+h) - \sin x \right]+ \left[\cos (x+h) - \cos x \right]}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[2 \cos \frac{2x+h}{2} \sin \frac{h}{2} \right]+ \left[-2 \sin \frac{2x+h}{2} \sin \frac{h}{2} \right]}{h} \right]}~\color{green}{\text{- - - I}}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{2 \sin \frac{h}{2} \left[\cos \frac{2x+h}{2}~-~\sin \frac{2x+h}{2} \right]}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin \frac{h}{2}}{h/2} \right]}~ × ~\lim_{h\rightarrow 0}{\left[\cos \frac{2x+h}{2}~-~\sin \frac{2x+h}{2} \right]}} &{} \\

{}&{}
& {~=~}& {1~ × ~\left[\cos \frac{2x+0}{2}~-~\sin \frac{2x+0}{2} \right]} &{} \\

{}&{}
& {~=~}& {1~ × ~\left[\cos x~-~\sin x \right]} &{} \\

{}&{}
& {~=~}& {\cos x~-~\sin x} &{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
Here we use the following identities:
    ♦ sin A - sin B = 2 cos (A+B)/2 sin (A-B)/2
    ♦ cos A - cos B = -2 sin (A+B)/2 sin (A-B)/2

Part (ii):
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{(x+h) \sin (x+h) ~-~x \sin x}{h}   \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x \sin (x+h) + h \sin (x+h) - x \sin x}{h}\right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x\left[\sin (x+h)  -  \sin x \right] + h \sin (x+h)}{h}   \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x\left[2 \cos \frac{2x+h}{2} \sin \frac{h}{2} \right] + h \sin (x+h)}{h} \right]}~\color{green}{\text{- - - I}}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{2x \cos \frac{2x+h}{2} \sin \frac{h}{2}}{h} \right]}~+~\lim_{h\rightarrow 0}{\left[\frac{h \sin (x+h)}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\left[\lim_{h\rightarrow 0}{\left[x \cos \frac{2x+h}{2} \right]}~ × ~\lim_{h\rightarrow 0}{\left[\frac{\sin \frac{h}{2}}{h/2} \right]} \right]

~+~\lim_{h\rightarrow 0}{\left[ \sin (x+h) \right]}} &{} \\

{}&{}
& {~=~}& {\left[{\left[x \cos \frac{2x+0}{2} \right]}~ × ~1 \right]

~+~{\left[\sin (x+0) \right]}} &{} \\

{}&{}
& {~=~}& {x \cos x + \sin x} &{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
Here we use the following identity:
    ♦ sin A - sin B = 2 cos (A+B)/2 sin (A-B)/2

Solved example 13.21
Compute the derivative of
(i) f(x) = sin 2x  (ii) g(x) = cot x
Solution:
Part (i):
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {(\sin 2x)'} &{} \\

{}&{}
& {~=~}& {(2 \sin x \cos x)'} &{} \\

{}&{}
& {~=~}& {(2 \sin x)' (\cos x)~+~(2\sin x)(\cos x)' \color{green}{\text{- - - I}}} &{} \\

{}&{}
& {~=~}& {(2 \sin x)' (\cos x)~+~(2\sin x)(-\sin x)} &{} \\

{}&{}
& {~=~}& {(2 \sin x)' (\cos x)~-~2 \sin^2 x} &{} \\

{}&{}
& {~=~}& {\left[(2)' \sin x ~+~ (2) (\sin x)' \right]\cos x~-~2 \sin^2 x} &{} \\

{}&{}
& {~=~}& {\left[0 × \sin x ~+~ (2) (\cos x) \right]\cos x~-~2 \sin^2 x} &{} \\

{}&{}
& {~=~}& {\left[2 \cos x \right]\cos x~-~2 \sin^2 x} &{} \\

{}&{}
& {~=~}& {2 \cos^2 x~-~2 \sin^2 x} &{} \\

{}&{}
& {~=~}& {2 (\cos^2 x - \sin^2 x)} &{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
Here we use the product rule: (uv)' = u'v + uv'

Part (ii):
• Here we want to find the derivative of cot x.
• This is already done before.
• See question number 11(v) of Exercise 13.2.

Solved example 13.22
Find the derivative of
(i) $\frac{x^5 - \cos x}{\sin x}$  (ii) $\frac{x + \cos x}{\tan x}$
Solution:
Part (i):
1. Here we can use quotient rule:
$\left(\frac{u}{v}\right)' = \frac{u' v - u v'}{u^2}$
• In our present case,
   ♦ u = x⁵ - cos x
   ♦ v = sin x
2. Thus we get:
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\frac{\left[(x^5 - \cos x)' \sin x \right] ~-~\left[(x^5 - \cos x) (\sin x)' \right]}{\sin^2 x}} &{} \\

{}&{}
& {~=~}& {\frac{\left[(5 x^4 + \sin x) \sin x \right] ~-~\left[(x^5 - \cos x) (\cos x) \right]}{\sin^2 x}} &{} \\

{}&{}
& {~=~}& {\frac{\left[5 x^4 \sin x + \sin^2 x \right] ~-~\left[x^5 \cos x - \cos^2 x \right]}{\sin^2 x}} &{} \\

{}&{}
& {~=~}& {\frac{5 x^4 \sin x + \sin^2 x  - x^5 \cos x + \cos^2 x}{\sin^2 x}} &{} \\

{}&{}
& {~=~}& {\frac{5 x^4 \sin x - x^5 \cos x + 1}{\sin^2 x}} &{} \\

{}&{}
& {~=~}& {\frac{- x^5 \cos x + 5 x^4 \sin x + 1}{\sin^2 x}} &{} \\

\end{array}$

Part (ii):
1. Here we can use quotient rule:
$\left(\frac{u}{v}\right)' = \frac{u' v - u v'}{u^2}$
• In our present case,
   ♦ u = x + cos x
   ♦ v = tan x
2. Thus we get:
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\frac{\left[(x + \cos x)' \tan x \right] ~-~\left[(x + \cos x) (\tan x)' \right]}{\tan^2 x}} &{} \\

{}&{}
& {~=~}& {\frac{\left[(1- \sin x) \tan x \right] ~-~\left[(x + \cos x) (\sec^2 x) \right]}{\tan^2 x}} &{} \\

\end{array}$

---

Link to a few more miscellaneous examples is given below:

Miscellaneous Exercise on chapter 13

---

In the next chapter, we will see mathematical reasoning.

Previous

Contents

Next

Copyright©2023 Higher secondary mathematics.blogspot.com

Chapter 13.17 - Derivatives of Trigonometric Functions

In the previous section, we saw the derivatives of polynomial functions. In this section, we will see the derivatives of trigonometric functions.

• We have already seen the derivative of sin x.
    ♦ That is., when f(x) = sin x, f'(x) will be cos x
• We obtained this result using first principle.
    ♦ That is., we used the formula:
$f'(x) = \lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}$
[See fig.13.33 in section 13.13]
• To find derivatives of various trigonometric functions, we need to know the derivative of cos x also.
• To find the derivative of cos x, we must use the first principle.
• Once we know the derivative of both sin x and cos x, we can use sum rule, product rule, quotient rule etc.,

• The derivative of cos x can be obtained as follows:
$\begin{array}{ll} {}&{f'(x)} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}} &{} \\

  {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[\cos (x+h)] – [cos x]}{h} \right]}} &{} \\

  {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[\cos x \cos h ~-~\sin x \sin h] – [cos x]}{h} \right]}} &{} \\

  {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\cos x (\cos h - 1)~-~ \sin x \sin h}{h} \right]}} &{} \\

  {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\cos x (\cos h - 1)}{h}
\right]}~-~\lim_{h\rightarrow 0}{\left[\frac{\sin x \sin h}{h} \right]}} &{} \\

  {}&{} & {~=~}& {\cos x \lim_{h\rightarrow 0}{\left[\frac{\cos h - 1}{h}
\right]}~-~\sin x \lim_{h\rightarrow 0}{\left[\frac{\sin h}{h} \right]}} &{} \\

{}&{} & {~=~}& {\cos x  × 0~-~\sin x  × 1} &{} \\
  {}&{} & {~=~}& {- \sin x} &{} \\ 

\end{array}$

• Now we have two useful results:
    ♦ If f(x) = sin x, then f'(x) = cos x  
    ♦ If f(x) = cos x, then f'(x) = -sin x

• We have already seen the graph of sin x and it's derivative. Now we will see the graph of cos x and it's derivative. It can be written in 5 steps:

1. We know that, f’(x) is a function. So we must  be able to plot f’(x).
2. In fig.13.39 below, both f(x) and f’(x) are plotted in the same graph.

Fig.13.39

3. The two graphs are plotted in different colors.
    ♦ The red curve represents f(x)
    ♦ The green curve represents f’(x)
4. Let us see the practical application of the graph. It can be written using two examples:
(i) Suppose that, we want the derivative of f(x) at $\left(x=\frac{\pi}{6}\right)$. That is., we want $f' \left(\frac{\pi}{6}\right)$
• For that, we mark point A on the red curve. Here, $\left(x=\frac{\pi}{6}\right)$
• Next we draw a vertical line through A. This vertical line meets the green curve at A'.
• $f' \left(\frac{\pi}{6}\right)$ will be equal to the y-coordinate of A'. In this case, it is $-\frac{1}{2}$
• We can verify this theoretically:
$f' \left(\frac{\pi}{6}\right)~=~- \sin \left(\frac{\pi}{6}\right)~=~- \frac{1}{2}$
(ii) Suppose that, we want the derivative of f(x) at $\left(x = -\pi \right)$. That is., we want $f' \left(-\pi \right)$
• For that, we mark point B on the red curve. Here, $\left(x = -\pi \right)$
• Next we draw a vertical line through B. This vertical line meets the green curve at B'.
• $f' \left(-\pi \right)$ will be equal to the y-coordinate of B'. In this case, it is 0.
• We can verify this theoretically:
$f' \left(-\pi \right)~=~- \sin \left(-\pi \right)~=~0$
5. Further more, the reader may verify the tangents also:
• Through the point A, draw a line at a slope of $- \frac{1}{2}$. This line will be tangential to f(x).
• Through the point B, draw a line at a slope of 0 (a horizontal line). This line will be tangential to f(x).   

---

Let us see some solved examples:

Solved example 13.17
Compute the derivative of tan x
Solution:
1. We can apply the quotient rule:
$\left(\frac{u}{v} \right)' = \frac{u'v - uv' }{v^2}$
2. In our present case:
u = sin x. So u' = cos x
v = cos x. So v' = -sin x
3. Thus we get:

$\begin{array}{ll}
{}&{\left(\frac{\sin x}{\cos x} \right)'}
& {~=~}& {\left(\frac{u}{v} \right)'}
&{} \\

{}&{}
& {~=~}& {\frac{[\cos x × \cos x] - [\sin x × -\sin x] }{\cos^2 x}}
&{} \\

{}&{}
& {~=~}& {\frac{\cos^2 x + \sin^2 x}{\cos^2 x}}
&{} \\

{}&{}
& {~=~}& {\frac{1}{\cos^2 x}}
&{} \\

{}&{}
& {~=~}& {\sec^2 x}
&{} \\

\end{array}$

Alternate method:
We can use first principle:

$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\tan(x+h) – \tan x}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{ \frac{\sin (x+h)}{\cos (x+h)} - \frac{\sin x}{\cos x}}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{ \frac{\sin (x+h) \cos x~-~\cos (x+h) \sin x}{\cos (x+h) \cos x}}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin (x+h) \cos x~-~\cos (x+h) \sin x}{h\;\cos (x+h) \cos x} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin (x+h - x)}{h\;\cos (x+h) \cos x} \right]}~\color{green}{\text{- - - (I)}}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin h}{h\;\cos (x+h) \cos x} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin h}{h} \right]}~ × ~\lim_{h\rightarrow 0}{\left[\frac{1}{\cos (x+h) \cos x} \right]}}
&{} \\

{}&{}
& {~=~}& {1~ × ~\frac{1}{\cos^2 x}}
&{} \\

{}&{}
& {~=~}& {\sec^2 x}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
Here we use the identity:
sin (A-B) = sin A cos B - cos A sin B

---

A graphical description can be written in 5 steps:
1. We know that, f’(x) is a function. So we must  be able to plot f’(x).
2. In fig.13.40 below, both f(x) and f’(x) are plotted in the same graph.

Fig.13.40

3. The two graphs are plotted in different colors.
    ♦ The red curve represents f(x)
    ♦ The green curve represents f’(x)
4. Let us see the practical application of the graph. It can be written using two examples:
(i) Suppose that, we want the derivative of f(x) at $\left(x=\frac{\pi}{3}\right)$. That is., we want $f' \left(\frac{\pi}{3}\right)$
• For that, we mark point A on the red curve. Here, $\left(x=\frac{\pi}{3}\right)$
• Next we draw a vertical line through A. This vertical line meets the green curve at A'.
• $f' \left(\frac{\pi}{3}\right)$ will be equal to the y-coordinate of A'. In this case, it is 4
• We can verify this theoretically:
$f' \left(\frac{\pi}{3}\right)~=~ \sec^2 \left(\frac{\pi}{3}\right)~=~\frac{1}{\cos^2 \left(\frac{\pi}{3}\right)}~=~\frac{1}{(1/2)^2}~=~4$
(ii) Suppose that, we want the derivative of f(x) at $\left(x = -\pi \right)$. That is., we want $f' \left(-\pi \right)$
• For that, we mark point B on the red curve. Here, $\left(x = -\pi \right)$
• Next we draw a vertical line through B. This vertical line meets the green curve at B'.
• $f' \left(-\pi \right)$ will be equal to the y-coordinate of B'. In this case, it is 1.
• We can verify this theoretically:
$f' \left(-\pi\right)~=~ \sec^2 \left(-\pi\right)~=~\frac{1}{\cos^2 \left(-\pi\right)}~=~\frac{1}{(-1)^2}~=~1$
5. Further more, the reader may verify the tangents also:
• Through the point A, draw a line at a slope of 4. This line will be tangential to f(x).
• Through the point B, draw a line at a slope of 1. This line will be tangential to f(x).

Solved example 13.18
Compute the derivative of f(x) = sin² x
Solution:
1. We can write sin² x as sin x . sin x
Now we can apply the product rule: (uv)' = u'v + uv'
2. The steps are shown below:
$\begin{array}{ll}
{}&{\left(\sin^2 x \right)'}
& {~=~}& {\left(\sin x \, . \, \sin x \right)'}
&{} \\

{}&{}
& {~=~}& {(\sin x)' × \sin x~+~\sin x  × (\sin x)'}
&{} \\

{}&{}
& {~=~}& {\cos x × \sin x~+~\sin x  × \cos x}
&{} \\

{}&{}
& {~=~}& {2 \sin x \cos x}
&{} \\

{}&{}
& {~=~}& {\sin 2x}
&{} \\

\end{array}$

A graphical description can be written in 5 steps:
1. We know that, f’(x) is a function. So we must  be able to plot f’(x).
2. In fig.13.41 below, both f(x) and f’(x) are plotted in the same graph.

Fig.13.41

3. The two graphs are plotted in different colors.
    ♦ The red curve represents f(x)
    ♦ The green curve represents f’(x)
4. Let us see the practical application of the graph. It can be written using two examples:
(i) Suppose that, we want the derivative of f(x) at $\left(x=\frac{\pi}{6}\right)$. That is., we want $f' \left(\frac{\pi}{6}\right)$
• For that, we mark point A on the red curve. Here, $\left(x=\frac{\pi}{6}\right)$
• Next we draw a vertical line through A. This vertical line meets the green curve at A'.
• $f' \left(\frac{\pi}{6}\right)$ will be equal to the y-coordinate of A'. In this case, it is 4
• We can verify this theoretically:
$f' \left(\frac{\pi}{6}\right)~=~ \sin \left(\frac{2 \pi}{6}\right)~=~ \sin \left(\frac{\pi}{3}\right) ~=~\frac{\sqrt3}{2}$
(ii) Suppose that, we want the derivative of f(x) at $\left(x=-\frac{\pi}{4}\right)$. That is., we want $f' \left(-\frac{\pi}{4}\right)$
• For that, we mark point B on the red curve. Here, $\left(x=-\frac{\pi}{4}\right)$
• Next we draw a vertical line through B. This vertical line meets the green curve at B'.
• $f' \left(-\frac{\pi}{4}\right)$ will be equal to the y-coordinate of B'. In this case, it is 4
• We can verify this theoretically:
$f' \left(-\frac{\pi}{4}\right)~=~ \sin \left(\frac{-2 \pi}{4}\right)~=~ \sin \left(-\frac{\pi}{2}\right)~=~ -\sin \left(\frac{\pi}{2}\right) ~=~-1$
5. Further more, the reader may verify the tangents also:
• Through the point A, draw a line at a slope of $\frac{\sqrt3}{2}$. This line will be tangential to f(x).
• Through the point B, draw a line at a slope of -1. This line will be tangential to f(x).

---

• We have drawn a large number of graphs. Those graphs were drawn to demonstrate the close relation between f(x) and f'(x).
• From now on wards, we will not draw any more graphs. We will concentrate only on the method for obtaining f'(x).
• However, the interested reader may continue to draw them. 

---

Link to a few more solved examples is given below:

Exercise 13.2

---

In the next section, we will see some miscellaneous examples.

Previous

Contents

Next

Copyright©2023 Higher secondary mathematics.blogspot.com