23.25 - Evaluation of Definite Integrals by Substitution

In the previous section, we saw how the fundamental theorem of calculus can be used to find the definite integral. We saw some solved examples also. In this section, we will see how the method of substitution can be used to speed up the process. We will demonstrate the method using a solved example.

Solved Example 23.88
Evaluate the definite integral $\small{\int_{1}^{2}{\left[5x^4 \sqrt{x^5 + 1} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{-1}^{1}{\left[5x^4 \sqrt{x^5 + 1} \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[5x^4 \sqrt{x^5 + 1} \right]dx}}$

(a) Put $\small{u = (x^5 + 1)~\Rightarrow \frac{du}{dx} = 5x^4 \Rightarrow 5x^4 dx = du}$

(b) $\small{F~=~\int{\left[5x^4 \sqrt{x^5 + 1} \right]dx}~=~\int{\left[\sqrt{u} \right]du}}$

$\small{~=~\frac{u^{3/2}}{3/2}~=~\frac{2}{3} \left(x^5 + 1 \right)^{\frac{3}{2}}}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(1)\,-\,F(-1)~=~\left[\frac{2}{3} \left(x^5 + 1 \right)^{\frac{3}{2}} \right]_{-1}^{1}}$

$\small{~=~\left[\frac{2}{3} \left(1^5 + 1 \right)^{\frac{3}{2}} \right]~-~\left[\frac{2}{3} \left((-1)^5 + 1 \right)^{\frac{3}{2}} \right]}$

$\small{~=~\left[\frac{2}{3} \left(2 \right)^{\frac{3}{2}} \right]~-~\left[\frac{2}{3} \left(0 \right)^{\frac{3}{2}} \right]}$

$\small{~=~\left[\frac{4 \sqrt 2}{3} \left(2 \right)^{\frac{3}{2}} \right]~-~\left[0 \right]~=~\frac{4 \sqrt 2}{3}}$

• The above two steps are already familiar to us. Let us see a method to speed up the process. For that, we will analyze the upper and lower limits.

• We wrote: $\small{u = x^5 + 1}$.
So when x gets closer and closer to -1, u gets closer and closer to zero.
Also, when x gets closer and closer to 1, u gets closer and closer to 2.

• So when using u as the variable, the lower and upper limits are zero and 2 respectively.

• We wrote: $\small{F~=~\int{\left[\sqrt{u} \right]du}~=~\frac{u^{3/2}}{3/2}~=~\frac{2}{3}u^{\frac{3}{2}}}$

• So we can write:

$\small{I~=~F(2)\,-\,F(0)~=~\left[\frac{2}{3}u^{\frac{3}{2}} \right]_{0}^{2}}$

$\small{~=~\left[\frac{2}{3} \left(2 \right)^{\frac{3}{2}} \right]~-~\left[\frac{2}{3} \left(0 \right)^{\frac{3}{2}} \right]}$

$\small{~=~\frac{4 \sqrt 2}{3}}$

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Let us see a few more solved examples:

Solved Example 23.89
Evaluate the definite integral $\small{\int_{0}^{1}{\left[\frac{x}{x^2 + 1} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{1}{\left[\frac{x}{x^2 + 1} \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\frac{x}{x^2+1} \right]dx}}$

Put $\small{u~=~x^2 + 1 \Rightarrow \frac{du}{dx}~=~2x \Rightarrow 2x\,dx~=~du}$

So we have: $\small{F~=~\int{\left[\frac{2x}{2(x^2+1)} \right]dx}~=~\int{\left[\frac{1}{2(u)} \right]du}~=~\frac{\log \left|u \right|}{2}}$

2. We wrote: $\small{u~=~x^2 + 1}$

   ♦ When x approach the lower limit zero, u approach 1

   ♦ When x approach the upper limit 1, u approach 2

3. So the given integral can be written as:

$\small{I~=~\int_{0}^{1}{\left[\frac{x}{x^2 + 1} \right]dx}~=~\int_1^2{\left[\frac{1}{2(u)} \right]du}}$

4. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(2)\,-\,F(1)~=~\left[\frac{\log \left|u \right|}{2} \right]_{1}^{2}}$

$\small{~=~\left[\frac{\log \left(2 \right)}{2} \right]~-~\left[\frac{\log \left(1 \right)}{2} \right]~=~\left[\frac{\log \left(2 \right)}{2} \right]~-~\left[0 \right]}$

$\small{~=~\frac{\log \left(2 \right)}{2} }$

Solved Example 23.90
Evaluate the definite integral $\small{\int_{0}^{\frac{\pi}{2}}{\left[\sqrt{\sin \phi} \cos^5 \phi \right]d \phi}}$
Solution:
1. Let $\small{I~=~\int_{0}^{\frac{\pi}{2}}{\left[\sqrt{\sin \phi} \cos^5 \phi \right]d \phi}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\sqrt{\sin \phi} \cos^5 \phi \right]d \phi}}$

$\small{\Rightarrow F~=~\int{\left[\sqrt{\sin \phi}\, \cos \phi\,\left(\cos^2 \phi \right)^2 \right]d \phi}}$

$\small{\Rightarrow F~=~\int{\left[\sqrt{\sin \phi}\, \cos \phi\,\left(1 - \sin^2 \phi \right)^2 \right]d \phi}}$

Put $\small{u~=~\sin \phi \Rightarrow \frac{du}{d\phi}~=~\cos \phi \Rightarrow \cos \phi\,d\phi~=~du}$

So we have: $\small{F~=~\int{\left[\sqrt{\sin \phi}\, \cos \phi\,\left(1 - \sin^2 \phi \right)^2 \right]d\phi}~=~\int{\left[\sqrt{u}\,\left(1 - u^2 \right)^2 \right]du}}$

$\small{~=~\int{\left[\sqrt{u}\,\left(1 - 2 u^2 +  u^4 \right) \right]du}~=~\int{\left[u^{1/2}~-~2 u^{5/2}~+~u^{9/2} \right]du}}$

$\small{~=~\frac{u^{3/2}}{3/2}~-~\frac{2 u^{7/2}}{7/2}~+~\frac{u^{11/2}}{11/2}~=~\frac{2 u^{3/2}}{3}~-~\frac{4 u^{7/2}}{7}~+~\frac{2 u^{11/2}}{11}}$

2. We wrote: $\small{u~=~\sin \phi}$

   ♦ When $\small{\phi}$ approach the lower limit zero, u approach 0

   ♦ When $\small{\phi}$ approach the upper limit $\small{\frac{\pi}{2}}$, u approach 1

3. So the given integral can be written as:

$\small{I~=~\int_{0}^{1}{\left[u^{1/2}~-~2 u^{5/2}~+~u^{9/2} \right]d \phi}}$

4. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(1)\,-\,F(0)~=~\left[\frac{2 u^{3/2}}{3}~-~\frac{4 u^{7/2}}{7}~+~\frac{2 u^{11/2}}{11} \right]_{0}^{1}}$

$\small{~=~\left[\frac{2 }{3}~-~\frac{4}{7}~+~\frac{2 }{11} \right]~-~\left[0 \right]}$

$\small{~=~\frac{64}{231} }$

Solved Example 23.91
Evaluate the definite integral $\small{\int_{0}^{1}{\left[\frac{\tan^{-1}x}{1+x^2} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{1}{\left[\frac{\tan^{-1}x}{1+x^2} \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\frac{\tan^{-1}x}{1+x^2} \right]dx}}$

Put $\small{u~=~\tan^{-1}x \Rightarrow \frac{du}{dx}~=~\frac{1}{1+x^2} \Rightarrow \frac{dx}{1+x^2}~=~du}$

So we have: $\small{F~=~\int{\left[u \right]du}~=~\frac{u^2}{2}}$

2. We wrote: $\small{u~=~\tan^{-1}x}$

   ♦ When $\small{x}$ approach the lower limit zero, u approach 0

   ♦ When $\small{x}$ approach the upper limit $\small{1}$, u approach $\small{\frac{\pi}{4}}$

3. So the given integral can be written as:

$\small{I~=~\int_{0}^{\frac{\pi}{4}}{\left[u \right]du}}$

4. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(\frac{\pi}{4})\,-\,F(0)~=~\left[\frac{u^2}{2} \right]_{0}^{\frac{\pi}{4}}}$

$\small{~=~\left[\frac{\pi^2 }{32} \right]~-~\left[0 \right]~=~\frac{\pi^2 }{32}}$

Solved Example 23.92
Evaluate the definite integral $\small{\int_{0}^{1}{\left[\sin^{-1}\left(\frac{2x}{1+x^2} \right) \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{1}{\left[\sin^{-1}\left(\frac{2x}{1+x^2} \right) \right]dx}}$  

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\sin^{-1}\left(\frac{2x}{1+x^2} \right) \right]dx}}$

Put $\small{x~=~\tan u}$

$\small{\Rightarrow \sin^{-1}\left(\frac{2x}{1+x^2} \right)~=~\sin^{-1}\left(\frac{2\tan u}{1+\tan^2 u} \right)~=~\sin^{-1}\left(\frac{2\tan u}{\sec^2 u} \right)}$

$\small{~=~\sin^{-1}\left(2 \sin u \cos u \right)~=~\sin^{-1}\left(\sin 2u \right)~=~2u ~=~2 \tan^{-1}x}$

So we have:

$\small{F~=~\int{\left[2 \tan^{-1}x \right]dx}~=~2 \int{\left[\tan^{-1}x \right]dx}~=~2\left(x \tan^{-1}x~+~\frac{\log \left|1+x^2 \right|}{2} \right)}$

(See solved example 23.54 of section 23.17)

$\small{~=~2x \tan^{-1}x~+~\log \left|1+x^2 \right|}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(1)\,-\,F(0)~=~\left[2x \tan^{-1}x~+~\log \left|1+x^2 \right| \right]_{0}^{1}}$

$\small{~=~\left[2(1) \tan^{-1}1~+~\log \left|1+1^2 \right| \right]~-~\left[2(0) \tan^{-1}(0)~+~\log \left|1+(0)^2 \right| \right]~=~\frac{\pi^2 }{32}}$

$\small{~=~\left[(2)\frac{\pi}{4}~+~\log \left|2 \right| \right]~-~\left[0 \right]~=~\frac{\pi}{2}~+~\log (2) }$

Solved Example 23.93
Evaluate the definite integral $\small{\int_{0}^{2}{\left[x \sqrt{x+2} \right]dx}}$ (Put x+2 =  t²)
Solution:
1. Let $\small{I~=~\int_{0}^{2}{\left[x \sqrt{x+2} \right]dx}}$

Put $\small{x+2} =  t^2\Rightarrow x= t^2 - 2\Rightarrow\frac{dx}{dt}~=~2t \Rightarrow dx = 2t\, dt$

   ♦ When x approaches zero, t approaches $\small{\sqrt 2}$

   ♦ When x approaches 2, t approaches 2

So we get: $\small{F ~=~ \int_{\sqrt 2}^{2}{\left[(t^2 - 2) \sqrt{t^2} \right]2t\,dt}~=~ \int_{\sqrt 2}^{2}{\left[(t^2 - 2) t \right]2t\,dt}}$

$\small{~=~ \int_{\sqrt 2}^{2}{\left[2t^4 - 4t^2 \right]dx}~=~\frac{2 t^5}{5}~-~\frac{4t^3}{3}}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(2)\,-\,F(\sqrt 2)~=~\left[\frac{2 t^5}{5}~-~\frac{4t^3}{3} \right]_{\sqrt 2}^{2}}$

$\small{~=~\left[\frac{2 (2)^5}{5}~-~\frac{4(2)^3}{3} \right]~-~\left[\frac{2 (\sqrt 2)^5}{5}~-~\frac{4(\sqrt 2)^3}{3} \right]}$

$\small{~=~\left[\frac{2^6}{5}~-~\frac{2^5}{3} \right]~-~\left[\frac{2^3 (\sqrt 2)}{5}~-~\frac{2^3(\sqrt 2)}{3} \right]}$

$\small{~=~\frac{2^6}{5}~-~\frac{2^5}{3} ~-~\frac{2^3 (\sqrt 2)}{5}~+~\frac{2^3(\sqrt 2)}{3} }$

$\small{~=~\frac{(3)2^6~-~(5)2^5~-~(3)2^3(\sqrt 2)~+~(5)2^3(\sqrt 2)}{15}}$

$\small{~=~\frac{(3)2^6~-~(5)2^5~+~(2)2^3(\sqrt 2)}{15}~=~\frac{(3)2^6~-~(5)2^5~+~2^4(\sqrt 2)}{15}}$

$\small{~=~\frac{2^5 \left[6~-~5 \right]~+~2^4(\sqrt 2)}{15}~=~\frac{2^5 ~+~2^4(\sqrt 2)}{15}~=~\frac{2^4 \left[2 ~+~\sqrt 2 \right]}{15}}$

$\small{~=~\frac{16 \sqrt 2 \left[\sqrt 2 ~+~1 \right]}{15}}$

Solved Example 23.94
Evaluate the definite integral $\small{\int_{0}^{2}{\left[\frac{1}{x+4 - x^2} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{2}{\left[\frac{1}{x+4 - x^2} \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\frac{1}{x+4 - x^2} \right]dx}}$

• In our present case, a = -1, b = 1 and c = 4

2. So we can calculate u and $\small{k^2}$:

• $\small{u\,=\,x\,+\,\frac{b}{2a}~=~x\,+\,\frac{1}{2(-1)}~=~(x-\frac{1}{2})}$

• $\small{\pm k^2\,=\,\frac{c}{a}\,-\,\frac{b^2}{4a^2}~=~\frac{4}{-1}\,-\,\frac{(1)^2}{4(-1)^2}~=~-4\,-\,\frac{1}{4}~=~\frac{-17}{4}}$

3. So we want:

$\small{\int{\left[\frac{dx}{x+4 - x^2} \right]}~=~\frac{1}{a}\int{\left[\frac{du}{u^2~\pm k^2} \right]}}$

$\small{~=~\frac{1}{-1}\int{\left[\frac{dx}{(x-\frac{1}{2})^2~-~\frac{17}{4}} \right]}~=~(-1)\int{\left[\frac{dx}{(x-\frac{1}{2})^2~-~\frac{17}{4}} \right]}}$

[Recall that, we put u = x + b/(2a). So du = dx]

4. That means, we want the definite integral:

$\small{~(-1)\int_0^2{\left[\frac{dx}{(x-\frac{1}{2})^2~-~\frac{17}{4}} \right]}}$

(i) Put t = (x−1/2). Then dt/dx = 1, which gives dt = dx

   ♦ Also, when x approaches zero, t approaches -(1/2)

   ♦ Similarly, when x approaches 2, t approaches (3/2)

• So we want:

$\small{~(-1)\int_0^2{\left[\frac{dx}{(x-\frac{1}{2})^2~-~\frac{17}{4}} \right]}~=~~(-1)\int_{-1/2}^{3/2}{\left[\frac{dt}{(t)^2~-~\left(\frac{\sqrt{17}}{2} \right)^2} \right]}}$

(ii) We have formula: $\small{\int{\left[\frac{dt}{t^2\,-\,m^2} \right]}\,=\, \frac{1}{2m} \log \left| \frac{t-m}{t+m}  \right|}$

• In our present case, m = $\small{\frac{\sqrt{17}}{2}}$

5. So we get:
$\small{(-1)\int{\left[\frac{dt}{(t)^2~-~\left(\frac{\sqrt{17}}{2} \right)^2}  \right]}\,=\, \frac{-1}{\sqrt{17}} \log \left| \frac{t-\frac{\sqrt{17}}{2}}{t+\frac{\sqrt{17}}{2}}  \right|}$

6. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(-1/2)\,-\,F(3/2)~=~\left[\frac{-1}{\sqrt{17}} \log \left| \frac{t-\frac{\sqrt{17}}{2}}{t+\frac{\sqrt{17}}{2}}  \right| \right]_{-1/2}^{3/2}}$

$\small{~=~\left[\frac{-1}{\sqrt{17}} \log \left| \frac{\frac{3}{2}-\frac{\sqrt{17}}{2}}{\frac{3}{2}+\frac{\sqrt{17}}{2}}  \right| \right]~-~\left[\frac{-1}{\sqrt{17}} \log \left| \frac{\frac{-1}{2}-\frac{\sqrt{17}}{2}}{\frac{-1}{2}+\frac{\sqrt{17}}{2}}  \right|\right]}$

$\small{~=~\left[\frac{-1}{\sqrt{17}} \log \left|\frac{3 - \sqrt{17}}{3 + \sqrt{17}} \right| \right]~+~\left[\frac{1}{\sqrt{17}} \log \left|\frac{-1 - \sqrt{17}}{-1 + \sqrt{17}}  \right|\right]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\left[\log \left|\frac{-1 - \sqrt{17}}{-1 + \sqrt{17}}  \right|~-~ \log \left|\frac{3 - \sqrt{17}}{3 + \sqrt{17}} \right| \right]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\left[\log \left|\frac{(-1)(1 + \sqrt{17})}{(-1)(1 - \sqrt{17})}  \right|~-~ \log \left|\frac{3 - \sqrt{17}}{3 + \sqrt{17}} \right| \right]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\left[\log \left|\frac{1 + \sqrt{17}}{1 - \sqrt{17}}  \right|~-~ \log \left|\frac{3 - \sqrt{17}}{3 + \sqrt{17}} \right| \right]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\big[\log\left[(1 + \sqrt{17})(3 + \sqrt{17}) \right]~-~\log\left[(1 - \sqrt{17})(3 - \sqrt{17}) \right] \big]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\big[\log\left[20 + 4 \sqrt {17} \right]~-~\log\left[20 - 4 \sqrt{17} \right] \big]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{20 + 4\sqrt{17}}{20 - 4\sqrt{17}} \right]  \bigg]~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{4(5 + \sqrt{17})}{4(5 - \sqrt{17})} \right]  \bigg]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{5 + \sqrt{17}}{5 - \sqrt{17}} \right]  \bigg]~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{(5 + \sqrt{17})(5 + \sqrt{17})}{(5 - \sqrt{17})(5 + \sqrt{17})} \right]  \bigg]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{25 + 10 \sqrt{17} + 17}{25 - 17} \right]  \bigg]~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{42 + 10 \sqrt{17}}{8} \right]  \bigg]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{21 + 5 \sqrt{17}}{4} \right]  \bigg]}$

Solved Example 23.95
Evaluate the definite integral $\small{\int_{-1}^{1}{\left[\frac{1}{x^2 + 2x + 5} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{-1}^{1}{\left[\frac{1}{x^2 + 2x + 5} \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\frac{1}{x^2 + 2x + 5} \right]dx}}$

• In our present case, a = 1, b = 2 and c = 5

2. So we can calculate u and $\small{k^2}$:

• $\small{u\,=\,x\,+\,\frac{b}{2a}~=~x\,+\,1}$

• $\small{\pm k^2\,=\,\frac{c}{a}\,-\,\frac{b^2}{4a^2}~=~\frac{5}{1}\,-\,\frac{(2)^2}{4(1)^2}~=~5\,-\,1~=~4}$

3. So we want:

$\small{\int{\left[\frac{dx}{x^2 + 2x + 5} \right]}~=~\frac{1}{a}\int{\left[\frac{du}{u^2~\pm k^2} \right]}}$

$\small{~=~\frac{1}{1}\int{\left[\frac{dx}{(x+1)^2~+~2^2} \right]}}$

[Recall that, we put u = x + b/(2a). So du = dx]

4. That means, we want the definite integral:

$\small{\int_{-1}^1{\left[\frac{dx}{(x+1)^2~+~2^2} \right]}}$

(i) Put t = (x+1). Then dt/dx = 1, which gives dt = dx

   ♦ Also, when x approaches -1, t approaches zero

   ♦ Similarly, when x approaches 1, t approaches 2

• So we want:

$\small{\int_{-1}^1{\left[\frac{dx}{(x+1)^2~+~2^2} \right]}~=~~\int_{0}^{2}{\left[\frac{dt}{(t)^2~+~\left(2 \right)^2} \right]}}$

(ii) We have formula: $\small{\int{\left[\frac{dt}{t^2\,+\,m^2} \right]}\,=\, \frac{1}{m} \tan^{-1}\left( \frac{t}{m} \right)}$

• In our present case, m = 2

5. So we get:
$\small{\int{\left[\frac{dt}{(t)^2~+~\left(2 \right)^2}  \right]}\,=\, \frac{1}{2} \tan^{-1}\left( \frac{t}{2} \right)}$

6. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(2)\,-\,F(0)~=~\left[\frac{1}{2} \tan^{-1}\left( \frac{t}{2} \right) \right]_{0}^{2}}$

$\small{~=~\left[\frac{1}{2} \tan^{-1}\left( \frac{2}{2} \right) \right]~-~\left[\frac{1}{2} \tan^{-1}\left( \frac{0}{2} \right)\right]}$

$\small{~=~\left[\frac{1}{2} \tan^{-1}\left( 1 \right) \right]~-~\left[\frac{1}{2} \tan^{-1}\left(0 \right)\right]}$

$\small{~=~\left[\frac{1}{2} \left(\frac{\pi}{4} \right) \right]~-~\left[\frac{1}{2} \left(0 \right)\right]}$

$\small{~=~\frac{\pi}{8}}$

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The link below gives a few more solved examples:

Exercise 23.10

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

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23.4 - Standard Integrals of Trigonometric Functions

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.

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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:
$\small{\int{\left[\tan x \right]dx~=~\log \left[|\sec x| \right]\,+\,C}}$

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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:
$\small{\int{\left[\cot x \right]dx~=~\log \left[|\sin x| \right]\,+\,C}}$

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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:
$\small{\int{\left[\sec x \right]dx~=~\log \left[|\sec x\,+\,\tan x| \right]\,+\,C}}$

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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:
$\small{\log \left[\left|\frac{1}{\csc x\,+\,\cot x} \right| \right]\,+\,C~=~\log \left[\left|\frac{\csc x\,-\,\cot x}{(\csc x\,+\,\cot x)(\csc x\,-\,\cot x)} \right| \right]\,+\,C}$

= $\small{\log \left[\left|\frac{\csc x\,-\,\cot x}{(\csc x\,+\,\cot x)(\csc x\,-\,\cot x)} \right| \right]\,+\,C~=~\log \left[\left|\frac{\csc x\,-\,\cot x}{\csc^2 x\,-\,\cot^2 x} \right| \right]\,+\,C}$

= $\small{\log \left[\left|\frac{\csc x\,-\,\cot x}{1} \right| \right]\,+\,C~=~\log\left[|\csc x\,-\,\cot x| \right]\,+\,C}$ 

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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. $\small{\int{\left[\csc x \right]dx}~=~\log\left[|\csc x\,-\,\cot x| \right]\,+\,C}$

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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}}$                           

• 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.

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

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