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:
$\small{I~=~F(2)\,-\,F(0)~=~\left[\frac{2}{3}u^{\frac{3}{2}} \right]_{0}^{2}}$
$\small{~=~\left[\frac{2}{3} \left(2 \right)^{\frac{3}{2}} \right]~-~\left[\frac{2}{3} \left(0 \right)^{\frac{3}{2}} \right]}$
$\small{~=~\frac{4 \sqrt 2}{3}}$
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:
$\small{I~=~F(2)\,-\,F(0)~=~\left[\frac{1}{2} \tan^{-1}\left( \frac{t}{2} \right) \right]_{0}^{2}}$
$\small{~=~\left[\frac{1}{2} \tan^{-1}\left( \frac{2}{2} \right) \right]~-~\left[\frac{1}{2} \tan^{-1}\left( \frac{0}{2} \right)\right]}$
$\small{~=~\left[\frac{1}{2} \tan^{-1}\left( 1 \right) \right]~-~\left[\frac{1}{2} \tan^{-1}\left(0 \right)\right]}$
$\small{~=~\left[\frac{1}{2} \left(\frac{\pi}{4} \right) \right]~-~\left[\frac{1}{2} \left(0 \right)\right]}$
$\small{~=~\frac{\pi}{8}}$
The link below gives a few more solved examples:
Exercise 23.10
In the next section, we will see a few more solved examples.
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