In the previous section, we saw integration by substitution. We saw some solved examples also. In this section, we will see some standard integrals of trigonometric functions.
In the table of integrals that we saw at the beginning of this chapter, there is the integral of sin x and also the integral of cos x. But altogether, there are six trigonometric functions. They are sine, cos, tan, csc, sec and cot. So there are four results missing from the table. This is because, integrals of sine and cosine can be obtained easily by inspection. The integrals of other four need to be derived using appropriate methods. We can use the method of substitution.
First we will find $\small{\int{\left[\tan x \right]dx}}$
1. We have: $\small{\tan x ~=~\frac{\sin x}{\cos x}}$
• −sin x is the derivative of cos x
• So we put u = cos x
⇒ $\small{\frac{du}{dx}}$ = −sin x
⇒ −sin x dx = du
2. So we want:
$\small{\int{\left[\tan x \right]dx}}$
= $\small{\int{\left[\frac{\sin x}{\cos x} \right]dx}~=~\int{\left[\frac{(-1) \sin x}{(-1) \cos x} \right]dx}}$
= $\small{\int{\left[\frac{1}{(-1) \cos x} \right]du}~=~\int{\left[\frac{1}{(-1) u} \right]du}~=~(-1)\int{\left[\frac{1}{ u} \right]du}}$
•
This integration gives:
$\small{- \log |u| \,+\,C~=~-\log[|u|] \,+\,C~=~\log[|u|]^{-1} \,+\,C~=~\log\left[\frac{1}{|u|} \right]\,+\,C}$
3. Substituting for u, we get:
$\small{\int{\left[\tan x \right]dx~=~\log\left[\frac{1}{|\cos x|} \right]\,+\,C~=~\log \left[|\sec x| \right]\,+\,C}}$
•
We can write:
$\small{\int{\left[\tan x \right]dx~=~\log \left[|\sec x| \right]\,+\,C}}$
Next we will find $\small{\int{\left[\cot x \right]dx}}$
1. We have: $\small{\cot x ~=~\frac{\cos x}{\sin x}}$
• cos x is the derivative of sin x
• So we put u = sin x
⇒ $\small{\frac{du}{dx}}$ = cos x
⇒ cos x dx = du
2. So we want:
$\small{\int{\left[\cot x \right]dx}}$
= $\small{\int{\left[\frac{\cos x}{\sin x} \right]dx}}$
=
$\small{\int{\left[\frac{1}{\sin x}
\right]du}~=~\int{\left[\frac{1}{ u} \right]du}}$
•
This integration gives:
$\small{\log[|u|] \,+\,C}$
3. Substituting for u, we get:
$\small{\int{\left[\cot x \right]dx~=~\log \left[|\sin x| \right]\,+\,C}}$
Next we will find $\small{\int{\left[\sec x \right]dx}}$
1. We have: $\small{\sec x ~=~\frac{\sec x(\sec x\,+\,\tan x)}{\sec x\,+\,\tan x}~=~\frac{\sec^2 x\,+\, \sec x \, \tan x}{\sec x\,+\,\tan x}}$
• (sec2 x + sec x tan x) is the derivative of (tan x + sec x)
• So we put u = tan x + sec x
⇒ $\small{\frac{du}{dx}~=~\sec^2 x\,+\, \sec x \, \tan x}$
⇒ (sec2 x + sec x tan x) dx = du
2. So we want:
$\small{\int{\left[\sec x \right]dx}}$
= $\small{\int{\left[\frac{\sec^2 x\,+\, \sec x \, \tan x}{\sec x\,+\,\tan x} \right]dx}}$
=
$\small{\int{\left[\frac{1}{\sec x\,+\,\tan x}
\right]du}~=~\int{\left[\frac{1}{ u} \right]du}}$
•
This integration gives:
$\small{\log[|u|] \,+\,C}$
3. Substituting for u, we get:
$\small{\int{\left[\sec x \right]dx~=~\log \left[|\sec x\,+\,\tan x| \right]\,+\,C}}$
Finally we will find $\small{\int{\left[\csc x \right]dx}}$
1.
We have: $\small{\csc x ~=~\frac{\csc x(\csc x\,+\,\cot x)}{\csc
x\,+\,\cot x}~=~\frac{\csc^2 x\,+\, \csc x \, \cot x}{\csc x\,+\,\cot
x}}$
• (−csc2 x − csc x cot x) is the derivative of (cot x + csc x)
• So we put u = cot x + csc x
⇒ $\small{\frac{du}{dx}~=~-\csc^2 x\,-\, \csc x \, \cot x}$
⇒ −(csc2 x + csc x cot x) dx = du
2. So we want:
$\small{\int{\left[\csc x \right]dx}}$
= $\small{\int{\left[\frac{(-1)(\csc^2 x\,+\, \csc x \, \cot x)}{(-1)(\csc x\,+\,\cot x)} \right]dx}}$
=
$\small{\int{\left[\frac{1}{(-1)(\csc x\,+\,\cot x)}
\right]du}~=~(-1) \int{\left[\frac{1}{u} \right]du}}$
•
This integration gives:
$\small{(-1) \log[|u|] \,+\,C~=~\log[|u|]^{-1} \,+\,C~=~ \log[\frac{1}{|u|}] \,+\,C}$
3. Substituting for u, we get:
$\small{\log \left[\left|\frac{1}{\csc x\,+\,\cot x} \right| \right]\,+\,C~=~\log \left[\left|\frac{\csc x\,-\,\cot x}{(\csc x\,+\,\cot x)(\csc x\,-\,\cot x)} \right| \right]\,+\,C}$
= $\small{\log \left[\left|\frac{\csc x\,-\,\cot x}{(\csc x\,+\,\cot x)(\csc x\,-\,\cot x)} \right| \right]\,+\,C~=~\log \left[\left|\frac{\csc x\,-\,\cot x}{\csc^2 x\,-\,\cot^2 x} \right| \right]\,+\,C}$
= $\small{\log \left[\left|\frac{\csc x\,-\,\cot x}{1} \right| \right]\,+\,C~=~\log\left[|\csc x\,-\,\cot x| \right]\,+\,C}$
Let us compile the six results related to trigonometric ratios:
1. $\small{\int{\left[\sin x \right]dx~=~- \cos x \,+\,C}}$
2. $\small{\int{\left[\cos x \right]dx~=~ \sin x \,+\,C}}$
3. $\small{\int{\left[\tan x \right]dx~=~\log \left[|\sec x| \right]\,+\,C}}$
4. $\small{\int{\left[\cot x \right]dx~=~\log \left[|\sin x| \right]\,+\,C}}$
5. $\small{\int{\left[\sec x \right]dx~=~\log \left[|\sec x\,+\,\tan x| \right]\,+\,C}}$
6. $\small{\int{\left[\csc x \right]dx}~=~\log\left[|\csc x\,-\,\cot x| \right]\,+\,C}$
Now we will see some solved examples:
Solved example 23.6
Find the following integrals:
(i) $\small{\int{\left[\sin^3 x \, \cos^2 x \right]dx}}$
(ii) $\small{\int{\left[\frac{\sin x}{\sin(x+a)} \right]dx}}$
(iii) $\small{\int{\left[\frac{1}{1\,+\, \tan x} \right]dx}}$
Solution:
Part (i):
1.
−sin x is the derivative of
cos x. There is no −ve sign in the question.
But it can be adjusted by multiplying two −ve signs.
• So we put u = cos x
⇒ $\small{\frac{du}{dx}~=~-\sin x}$
⇒ −sinx dx = du
•
Also, when u = cos x, we get: sin2x = 1 − u2
2. So we want:
$\small{\int{\left[\sin^3 x \, \cos^2 x \right]dx}~=~\int{\left[(-1)(-1) \sin^3 x \, \cos^2 x \right]dx}}$
= $\small{\int{\left[(-1)\sin^2 x \,\, u^2 \right]du}~=~\int{\left[(-1)(1 - u^2) \, u^2 \right]du}}$
•
This integration can be done as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[(-1)(1 - u^2) \, u^2\right] \, du}} & {~=~} &{\int{\left[-u^2 + u^4 \, \right] \, du}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\int{\left[-u^2 \, \right] \, du}~+~\int{\left[u^4 \, \right] \, du}} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\left[\frac{- u^{3}}{3}\,+\,C_1 \right]~+~\left[\frac{u^{5}}{5} \,+\,C_2 \right]} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{- u^{3}}{3}\,+\, \frac{u^{5}}{5} \,+\, C} \\
\end{array}}$
3. Substituting for u, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[\sin^3 x \, \cos^2 x\right] \, du}} & {~=~} &{\frac{-u^{3}}{3}\,+\, \frac{u^{5}}{5} \,+\, C} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\frac{-\cos^{3} x}{3}\,+\, \frac{\cos^{5} x}{5} \,+\, C} \\
\end{array}}$
Part (ii):
1.
The derivative of
(x+a) is 1.
• So we put u = x+a
⇒ $\small{\frac{du}{dx}~=~1}$
⇒ dx = du
•
Also, since u = x+a, we get: x = u−a
2. So we want:
$\small{\int{\left[\frac{\sin x}{\sin(x+a)} \right]dx}~=~\int{\left[\frac{\sin x}{\sin(u)} \right]dx}}$
= $\small{\int{\left[\frac{\sin (u-a)}{\sin(u)} \right]du}}$
•
This integration can be done as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[\frac{\sin x}{\sin (x+a)}\right] \, dx}} & {~=~} &{\int{\left[\frac{\sin (u-a)}{\sin u}\right] \, du}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\int{\left[\frac{\sin u \cos a \,-\, \cos u \sin a }{\sin u}\right] \, du}} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\int{\left[\cos a \,-\, \cot u \sin a\right] \, du}} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\int{\left[\cos a \right] \, du}~-~\int{\left[\cot u \sin a\right] \, du}} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\cos a \int{\left[1 \right] \, du}~-~ \sin a \int{\left[\cot u \right] \, du}} \\
{~\color{magenta} 6 } &{{}} &{{}} & {~=~} &{\cos a \left[u \,+\,C_1 \right]~-~\sin a \left[\log |\sin u | \,+\,C_2 \right]} \\
{~\color{magenta} 7 } &{{}} &{{}} & {~=~} &{(\cos a)u \,+\,(\cos a)C_1 ~-~(\sin a)\log |\sin u | \,-\,(\sin a) C_2} \\
{~\color{magenta} 8 } &{{}} &{{}} & {~=~} &{(\cos a)(x+a) \,+\,(\cos a)C_1 ~-~(\sin a)\log |\sin (x+a) | \,-\,(\sin a) C_2} \\
{~\color{magenta} 9 } &{{}} &{{}} & {~=~} &{x \cos a \,+\, a \cos a \,+\,(\cos a)C_1 ~-~(\sin a)\log |\sin (x+a) | \,-\,(\sin a) C_2} \\
{~\color{magenta} {10} } &{{}} &{{}} & {~=~} &{x \cos a \,+\, C_3 \,+\,C_4 ~-~(\sin a)\log |\sin (x+a) | \,-\,C_5} \\
{~\color{magenta} {11} } &{{}} &{{}} & {~=~} &{x \cos a \,-\, (\sin a)\log |\sin (x+a) | \,+\,C} \\
\end{array}}$
• Note that, the constants C1, C2, C3 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:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{1}{1\,+\,\tan x}} & {~=~} &{\frac{1}{1\,+\,(\sin x / \cos x)}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\frac{\cos x}{\sin x\,+\,\cos x}} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{\cos x}{\sin x\,+\,\sin(\pi/2 \,-\, x)}} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{\cos x}{2 \sin(\pi/4) \cos(x\,-\, \pi/4)}} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\frac{\cos x}{\sqrt{2} \cos(x\,-\, \pi/4)}} \\
\end{array}}$
•
Derivative of
(x−π/4) w.r.t x is 1.
• So we put u = (x−π/4)
⇒ $\small{\frac{du}{dx}~=~1}$
⇒ dx = du
•
Also, since u = (x−π/4), we get: x = u + π/4
2. So we want:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[\frac{\cos x}{\sqrt{2} \cos(x\,-\, \pi/4)}\right] \, dx}} & {~=~} &{\int{\left[\frac{\cos (u\,+\, \pi/4)}{\sqrt{2} \cos u}\right] \, du}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\int{\left[\frac{\cos u \cos (π/4)~-~\sin u \sin(π/4)}{\sqrt{2} \cos u}\right] \, du}} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\int{\left[\frac{\cos u (1/\sqrt 2)~-~\sin u (1/\sqrt 2)}{\sqrt{2} \cos u}\right] \, du}} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\int{\left[\frac{(1/\sqrt 2)~-~\tan u (1/\sqrt 2)}{\sqrt{2}}\right] \, du}} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\int{\left[\frac{(1/\sqrt 2)(1~-~\tan u)}{\sqrt{2}}\right] \, du}} \\
{~\color{magenta} 6 } &{{}} &{{}} & {~=~} &{\int{\left[\frac{1~-~\tan u}{2}\right] \, du}} \\
\end{array}}$
•
This integration can be done as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[\frac{1~-~\tan u}{2}\right] \, dx}} & {~=~} &{\int{\left[\frac{1}{2}\right] \, du}~-~\int{\left[\frac{\tan u}{2}\right] \, du}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\frac{1}{2} \int{\left[1 \right] \, du}~-~\frac{1}{2} \int{\left[\tan u \right] \, du}} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{1}{2} \left[u\,+\,C_1 \right]~-~\frac{1}{2} \left[\log |\sec u |\,+\, C_2 \right]} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{1}{2} \left[x\,-\,(\pi/4)\,+\,C_1 \right]~-~\frac{1}{2} \left[\log |\sec (x\,-\,(\pi/4)) |\,+\, C_2 \right]} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[- \log |\sec (x\,-\,(\pi/4)) |\,-\, C_2 \right]} \\
{~\color{magenta} 6 } &{{}} &{{}} & {~=~} &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[- \log \left|\frac{1}{\cos (x\,-\,(\pi/4))} \right |\,-\, C_2 \right]} \\
{~\color{magenta} 7 } &{{}} &{{}} & {~=~} &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log \left|\left(\frac{1}{\cos (x\,-\,(\pi/4))}\right)^{-1} \right |\,+\, C_4 \right]} \\
{~\color{magenta} 8 } &{{}} &{{}} & {~=~} &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log \left|\cos (x\,-\,(\pi/4)) \right |\,+\, C_4 \right]} \\
{~\color{magenta} 9 } &{{}} &{{}} & {~=~} &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log \left|\cos x \, \cos(\pi/4) ~+~\sin x \,\sin(\pi/4) \right |\,+\, C_4 \right]} \\
{~\color{magenta} {10} } &{{}} &{{}} & {~=~} &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log \left|\cos x \, (1/\sqrt 2) ~+~\sin x \,(1/\sqrt 2) \right |\,+\, C_4 \right]} \\
{~\color{magenta} {11} } &{{}} &{{}} & {~=~} &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log \left|(1/\sqrt 2)(\cos x ~+~\sin x) \right |\,+\, C_4 \right]} \\
{~\color{magenta} {12} } &{{}} &{{}} & {~=~} &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log (1/\sqrt 2)~+~ \log \left|(\cos x ~+~\sin x) \right |\,+\, C_4 \right]} \\
{~\color{magenta} {13} } &{{}} &{{}} & {~=~} &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log \left|(\cos x ~+~\sin x) \right |\,+\, C_5 \right]} \\
{~\color{magenta} {14} } &{{}} &{{}} & {~=~} &{\frac{x}{2} \,+\, \frac{1}{2} \left[\log \left|(\cos x ~+~\sin x) \right | \right]\,+\, C} \\
\end{array}}$
In the next section, we will see a few more solved examples.
Previous
Contents
Next
Copyright©2025 Higher secondary mathematics.blogspot.com
No comments:
Post a Comment