In the previous section, we saw the basic details about Integration by parts. We saw some solved examples also. In this section, we will see a few more solved examples.
Solved Example 23.45
Find $\small{\int{\left[x \, \log 2x \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = log 2x
♦ Let second function be: g(x) = x
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x\right]dx}~=~\frac{x^2}{2}}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\log 2x \, \left(\frac{x^2}{2} \right) \big]}$
• This is the first term.
4. $\small{f'(x)~=~\frac{1}{x}}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[\frac{1}{x}\,\left(\frac{x^2}{2} \right) \big]dx}~=~\frac{x^2}{4}}$
• This is the second term.
6. So we get:
$\small{\int{\left[x \, \log 2x \right]dx}~=~\text{First term - Second term}~=~\log 2x \, \left(\frac{x^2}{2} \right)~-~\frac{x^2}{4}~+~\rm{C}}$
Solved Example 23.46
Find $\small{\int{\left[x^2 \, \log x \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = log x
♦ Let second function be: g(x) = $\small{x^2}$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x^2\right]dx}~=~\frac{x^3}{3}}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\log x \, \left(\frac{x^3}{3} \right) \big]}$
• This is the first term.
4. $\small{~f'(x)~=~\frac{1}{x}}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[\frac{1}{x}\,\left(\frac{x^3}{3} \right) \big]dx}~=~\frac{x^3}{9}}$
• This is the second term.
6. So we get:
$\small{\int{\left[x^2 \, \log x \right]dx}~=~\text{First term - Second term}~=~\log x \, \left(\frac{x^3}{3} \right)~-~\frac{x^3}{9}~+~\rm{C}}$
Solved Example 23.47
Find $\small{\int{\left[x \, \sin^{-1} x \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = $\small{\sin^{-1} x}$
♦ Let second function be: g(x) = $\small{x}$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x \right]dx}~=~\frac{x^2}{2}}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sin^{-1}x \, \left(\frac{x^2}{2}\right) \big]}$
• This is the first term.
4. $\small{f'(x)~=~\frac{1}{\sqrt{1-x^2}}}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[\frac{1}{\sqrt{1-x^2}}\,\left(\frac{x^2}{2} \right) \big]dx}~=~\frac{1}{2}\int{\big[\frac{x^2}{\sqrt{1-x^2}} \big]dx}}$
• This is the second term. But is also a product. We will do it as a sub group:
*** Beginning of sub group ***
We want $\small{\frac{1}{2}\int{\big[\frac{x^2}{\sqrt{1-x^2}} \big]dx}}$
(i) Let x = sin u
$\small{\Rightarrow \frac{dx}{du}\,=\,\cos u \Rightarrow\frac{dx}{\cos u}\,=\,du}$
Also, $\small{~\sqrt{1 - x^2}\,=\,\cos u}$
(ii) So we want:
$\small{\frac{1}{2}\int{\big[\frac{\sin^2 u}{\cos u} \big]dx}\,=\,\frac{1}{2}\int{\big[\sin^2 u \big]du}}$
(iii) This can be calculated as:
$\small{\frac{1}{2}\int{\big[\sin^2 u \big]du}\,=\,\frac{1}{2}\int{\big[\frac{1 - \cos 2u}{2} \big]du}\,=\,\frac{1}{4}\int{\big[1 - \cos 2u \big]du}}$
$\small{~=~\frac{u}{4}\,-\,\frac{\sin 2u}{8}~=~\frac{u}{4}\,-\,\frac{2 \sin u \cos u}{8}}$
$\small{~=~\frac{u}{4}\,-\,\frac{\sin u \cos u}{4}~=~\frac{\sin^{-1}x}{4}\,-\,\frac{x \sqrt{1 - x^2}}{4}}$
*** End of sub group ***
6. So we get:
$\small{\int{\left[x \, \sin^{-1} x \right]dx}~=~\text{First term - Second term}}$
$\small{~=~\big[\sin^{-1}x \, \left(\frac{x^2}{2}\right) \big]~-~\big[\frac{\sin^{-1}x}{4}\,-\,\frac{x \sqrt{1 - x^2}}{4} \big]}$
$\small{~=~\sin^{-1}x \, \left(\frac{x^2}{2}\,-\,\frac{1}{4}\right)\,+\,\frac{x \sqrt{1 - x^2}}{4}\,+\,\rm{C}}$
Solved Example 23.48
Find $\small{\int{\left[x \, \tan^{-1} x \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = $\small{\tan^{-1} x}$
♦ Let second function be: g(x) = $\small{x}$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x \right]dx}~=~\frac{x^2}{2}}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\tan^{-1}x \, \left(\frac{x^2}{2}\right) \big]}$
• This is the first term.
4. $\small{f'(x)~=~\frac{1}{1+x^2}}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[\frac{1}{1+x^2}\,\left(\frac{x^2}{2} \right) \big]dx}~=~\frac{1}{2}\int{\big[\frac{x^2}{1+x^2} \big]dx}}$
• This is the second term. But is also a product. We will do it as a sub group:
*** Beginning of sub group ***
We want $\small{\frac{1}{2}\int{\big[\frac{x^2}{1+x^2} \big]dx}}$
(i) Let x = tan u
$\small{\Rightarrow \frac{dx}{du}\,=\,\sec^2 u \Rightarrow\frac{dx}{\sec^2 u}\,=\,du}$
Also, $\small{~1 + x^2\,=\,\sec^2 u}$
(ii) So we want:
$\small{\frac{1}{2}\int{\big[\frac{\tan^2 u}{\sec^2 u} \big]dx}\,=\,\frac{1}{2}\int{\big[\tan^2 u \big]du}}$
(iii) This can be calculated as:
$\small{\frac{1}{2}\int{\big[\tan^2 u \big]du}\,=\,\frac{1}{2}\int{\big[\sec^2 u \,-\,1 \big]du}\,=\,\frac{\tan u}{2}\,-\,\frac{u}{2}}$
$\small{\,=\,\frac{x}{2}\,-\,\frac{\tan^{-1}x}{2}}$
*** End of sub group ***
6. So we get:
$\small{\int{\left[x \, \tan^{-1} x \right]dx}~=~\text{First term - Second term}}$
$\small{~=~\big[\tan^{-1}x \, \left(\frac{x^2}{2}\right) \big]~-~\big[\frac{x}{2}\,-\,\frac{\tan^{-1}x}{2}\big]}$
$\small{~=~\tan^{-1}x \, \left(\frac{x^2}{2}\,+\,\frac{1}{2}\right)\,-\,\frac{x}{2}\,+\,\rm{C}}$
Solved Example 23.49
Find $\small{\int{\left[x \, \cos^{-1} x \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = $\small{\cos^{-1} x}$
♦ Let second function be: g(x) = $\small{x}$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x \right]dx}~=~\frac{x^2}{2}}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\cos^{-1}x \, \left(\frac{x^2}{2}\right) \big]}$
• This is the first term.
4. $\small{f'(x)~=~\frac{-1}{\sqrt{1-x^2}}}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[\frac{-1}{\sqrt{1-x^2}}\,\left(\frac{x^2}{2} \right) \big]dx}~=~\frac{-1}{2}\int{\big[\frac{1}{\sqrt{1-x^2}} \big]dx}}$
• This is the second term. But is also a product. We will do it as a sub group:
*** Beginning of sub group ***
We want $\small{\frac{-1}{2}\int{\big[\frac{x^2}{\sqrt{1-x^2}} \big]dx}}$
(i) Let x = cos u
$\small{\Rightarrow \frac{dx}{du}\,=\,-\sin u \Rightarrow\frac{-dx}{\sin u}\,=\,du}$
Also, $\small{~\sqrt{1 - x^2}\,=\,\sin u}$
(ii) So we want:
$\small{\frac{-1}{2}\int{\big[\frac{\cos^2 u}{\sin u} \big]dx}\,=\,\frac{1}{2}\int{\big[\cos^2 u \big]du}}$
(iii) This can be calculated as:
$\small{\frac{1}{2}\int{\big[\cos^2 u \big]du}\,=\,\frac{1}{2}\int{\big[\frac{1 + \cos 2u}{2} \big]du}\,=\,\frac{1}{4}\int{\big[1 + \cos 2u \big]du}}$
$\small{~=~\frac{u}{4}\,+\,\frac{\sin 2u}{8}~=~\frac{u}{4}\,+\,\frac{2 \sin u \cos u}{8}}$
$\small{~=~\frac{u}{4}\,+\,\frac{\sin u \cos u}{4}~=~\frac{\cos^{-1}x}{4}\,+\,\frac{x \sqrt{1 - x^2}}{4}}$
*** End of sub group ***
6. So we get:
$\small{\int{\left[x \, \cos^{-1} x \right]dx}~=~\text{First term - Second term}}$
$\small{~=~\big[\cos^{-1}x \, \left(\frac{x^2}{2}\right) \big]~-~\big[\frac{\cos^{-1}x}{4}\,+\,\frac{x \sqrt{1 - x^2}}{4} \big]}$
$\small{~=~\cos^{-1}x \, \left(\frac{x^2}{2}\,-\,\frac{1}{4}\right)\,-\,\frac{x \sqrt{1 - x^2}}{4}\,+\,\rm{C}}$
Solved Example 23.50
Find $\small{\int{\left[\sin^{-1}x \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = $\small{\sin^{-1} x}$
♦ Let second function be: g(x) = $\small{1}$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~x}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sin^{-1}x \, \left(x\right) \big]}$
• This is the first term.
4. $\small{f'(x)~=~\frac{1}{\sqrt{1-x^2}}}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[\frac{1}{\sqrt{1-x^2}}\,\left(x \right) \big]dx}~=~-\sqrt{1-x^2}}$
• This is second term.
(The reader may write all steps involved in this integration process)
6. So we get:
$\small{\int{\left[\sin^{-1} x \right]dx}~=~\text{First term - Second term}}$
$\small{~=~\big[\sin^{-1}x \, \left(x\right) \big]~-~\big[-\sqrt{1-x^2} \big]}$
$\small{~=~x \sin^{-1}x~+~\sqrt{1-x^2}\,+\,\rm{C}}$
Solved Example 23.51
Find $\small{\int{\left[(\sin^{-1}x)^2 \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = $\small{\sin^{-1} x}$
♦ Let second function be: g(x) = $\small{\sin^{-1}x}$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\sin^{-1}x \right]dx}~=~x \sin^{-1}x~+~\sqrt{1-x^2}}$
(See solved example 23.50 above)
3. $\small{\big[f(x) \left(A \right) \big]~=~\sin^{-1}x \big[x \sin^{-1}x~+~\sqrt{1-x^2} \big]}$
$\small{~=~\big[x (\sin^{-1}x)^2~+~\sin^{-1}x\, \left(\sqrt{1-x^2}\right) \big]}$
• This is the first term.
4. $\small{f'(x)~=~\frac{1}{\sqrt{1-x^2}}}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[\frac{1}{\sqrt{1-x^2}}\,\left(x \sin^{-1}x~+~\sqrt{1-x^2} \right) \big]dx}}$
$\small{~=~\int{\big[\frac{x \sin^{-1}x}{\sqrt{1-x^2}}~+~1 \big]dx}~=~x\,-\,(\sqrt{1-x^2}) \sin^{-1}x~+~x}$
$\small{~=~2x\,-\,\left(\sqrt{1-x^2} \right) \sin^{-1}x}$
(See solved example 23.38 of the previous section)
• This is second term.
6. So we get:
$\small{\int{\left[(\sin^{-1} x)^2 \right]dx}~=~\text{First term - Second term}}$
$\small{~=~\big[x (\sin^{-1}x)^2~+~\sin^{-1}x\, \left(\sqrt{1-x^2}\right) \big]~-~\big[2x\,-\,\left(\sqrt{1-x^2} \right) \sin^{-1}x \big]}$
$\small{~=~x (\sin^{-1}x)^2~+~2 \sin^{-1}x\, \left(\sqrt{1-x^2}\right) ~-~2x}$
Solved Example 23.52
Find $\small{\int{\left[\frac{x \cos^{-1} x}{\sqrt{1 - x^2}} \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = $\small{\cos^{-1}x}$
♦ Let second function be: g(x) = $\small{\frac{x }{\sqrt{1 - x^2}} }$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\frac{x }{\sqrt{1 - x^2}} \right]dx}~=~-\sqrt{1-x^2}}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[(-1)(\cos^{-1}x) \,\sqrt{1-x^2} \big]}$
• This is the first term.
4. $\small{f'(x)~=~\frac{-1}{\sqrt{1 - x^2}}}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[\frac{-1}{\sqrt{1 - x^2}}\,\left(-\sqrt{1-x^2} \right) \big]dx}~=~x}$
• This is the second term.
6. So we get:
$\small{\int{\left[\frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \right]dx}~=~\text{First term - Second term}}$
$\small{~=~\big[(-1)(\cos^{-1}x) \,\sqrt{1-x^2} \big]~-~\big[x \big]~+~\rm{C}}$
$\small{~=~- \big[x~+~(\cos^{-1}x) \,\sqrt{1-x^2}\big]~+~\rm{C}}$
Solved Example 23.53
Find $\small{\int{\left[x \sec^2 x \right]dx}}$
Solution:
$\small{\int{\left[x \sec^2 x \right]dx}}$
1. Assigning first and second functions:
♦ Let first function be: f(x) = $\small{x}$
♦ Let second function be: g(x) = $\small{\sec^2 x }$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\sec^2 x \right]dx}~=~\tan x}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[x \tan x \big]}$
• This is the first term.
4. $\small{f'(x)~=~1}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[1\,\left(\tan x \right) \big]dx}~=~\log \left|\sec x \right| }$
• This is the second term.
6. So we get:
$\small{\int{\left[\frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \right]dx}~=~\text{First term - Second term}}$
$\small{~=~\big[x \tan x \big]~-~\big[\log \left|\sec x \right| \big]~+~\rm{C}}$
$\small{~=~x \tan x ~+~\log \left|\frac{1}{\sec x} \right| ~+~\rm{C}}$
$\small{~=~x \tan x ~+~\log \left|\cos x \right| ~+~\rm{C}}$
Solved Example 23.54
Find $\small{\int{\left[\tan^{-1}x \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = $\small{\tan^{-1} x}$
♦ Let second function be: g(x) = $\small{1}$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~x}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\tan^{-1}x \, \left(x\right) \big]}$
• This is the first term.
4. $\small{f'(x)~=~\frac{1}{1+x^2}}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[\frac{1}{1+x^2}\,\left(x \right) \big]dx}~=~\frac{1}{2}\log \left|1+x^2 \right|}$
• This is second term.
(The reader may write all steps involved in this integration process)
6. So we get:
$\small{\int{\left[\tan^{-1} x \right]dx}~=~\text{First term - Second term}}$
$\small{~=~\big[\tan^{-1}x \, \left(x\right) \big]~-~\big[\frac{1}{2}\log \left|1+x^2 \right| \big]}$
$\small{~=~x \tan^{-1}x~-~\frac{1}{2}\log \left|1+x^2 \right|\,+\,\rm{C}}$
Solved Example 23.55
Find $\small{\int{\left[x (\log x)^2 \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = $\small{(\log x)^2}$
♦ Let second function be: g(x) = $\small{x}$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x \right]dx}~=~\frac{x^2}{2}}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[(\log x)^2 \, \left(\frac{x^2}{2}\right) \big]}$
• This is the first term.
4. $\small{f'(x)~=~\frac{2 \log \left|x \right|}{x}}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[\frac{2 \log \left|x \right|}{x}\,\left(\frac{x^2}{2} \right) \big]dx}}$
$\small{~=~\int{\big[x\log \left|x \right| \big]dx}~=~\log x \, \left(\frac{x^2}{2} \right)~-~\frac{x^2}{4}}$
(See solved example 23.44 of the previous section)
• This is second term.
6. So we get:
$\small{\int{\left[x (\log x)^2 \right]dx}~=~\text{First term - Second term}}$
$\small{~=~\big[(\log x)^2 \, \left(\frac{x^2}{2}\right) \big]~-~\big[\log x \, \left(\frac{x^2}{2} \right)~-~\frac{x^2}{4} \big]}$
$\small{~=~(\log x)^2 \, \left(\frac{x^2}{2}\right) ~-~\log x \, \left(\frac{x^2}{2} \right)~+~\frac{x^2}{4} }$
Solved Example 23.56
Find $\small{\int{\left[(x^2 + 1) \log x \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = log x
♦ Let second function be: g(x) = $\small{x^2 + 1}$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x^2 + 1\right]dx}~=~\frac{x^3}{3}~+~x}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\log x \, \left(\frac{x^3}{3}~+~x \right) \big]}$
• This is the first term.
4. $\small{f'(x)~=~\frac{1}{x}}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[\frac{1}{x}\,\left(\frac{x^3}{3}~+~x \right) \big]dx}~=~\frac{x^3}{9}~+~x}$
• This is the second term.
6. So we get:
$\small{\int{\left[(x^2 + 1) \log x \right]dx}~=~\text{First term - Second term}}$
$\small{~=~\log x \, \left(\frac{x^3}{3}~+~x \right)~-~\frac{x^3}{9}~-~x~+~\rm{C}}$
We
have seen the method of integration by parts.
In the next section, we will see a special case in this method.
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