In the previous section, we completed a discussion on integration by partial fractions. In this section, we will see the third method for integration, which is: Integration by parts.
This method is useful for integrating "products of functions". It can be explained in 6 steps:
1. Consider the product rule:
$\small{\frac{d}{dx}(uv)~=~u \frac{dv}{dx}\,+\,v \frac{du}{dx}}$
2. Integrating both sides, we get:
$\small{uv~=~\int{\left[u \frac{dv}{dx} \right]dx}\,+\,\int{\left[v \frac{du}{dx} \right]dx}}$
⇒ $\small{\int{\left[u \frac{dv}{dx} \right]dx}~=~uv\,-\,\int{\left[v \frac{du}{dx} \right]dx}}$
Note: In the above step, there will be a constant of integration. But as we continue to write more steps, there will be more constants. We just need to combine all those constants into a single constant. This can be done in the final step.
3. Let $\small{u = f(x) ~~\rm{and}~~\frac{dv}{dx} = g(x)}$
4. Then we can write:
$\small{\frac{du}{dx} = f'(x) ~~\rm{and}~~v = \int{\left[g(x) \right]dx}}$
5. Now the result in (2) becomes:
$\small{\int{\left[f(x)\, g(x) \right]dx}~=~\big[f(x) \left(\int{\left[g(x) \right]dx} \right) \big]\,-\,\int{\big[\left(\int{\left[g(x) \right]dx} \right)\,f'(x) \big]dx}}$
⇒ $\small{\int{\left[f(x)\, g(x) \right]dx}~=~\big[f(x) \left(\int{\left[g(x) \right]dx} \right) \big]\,-\,\int{\big[f'(x)\,\left(\int{\left[g(x) \right]dx} \right)\, \big]dx}}$
⇒ $\small{\int{\left[f(x)\, g(x) \right]dx}~=~\big[f(x) \left(A \right) \big]\,-\,\int{\big[f'(x)\,\left(A \right)\, \big]dx}~+~\rm{C}}$
Where $\small{A~=~\int{\left[g(x) \right]dx} }$
6. Let us compile the above steps:
• We are asked to integrate a function, which is the product of two functions.
• Let f(x) be the first function and g(x) be the second function.
The compiled form can be written in three steps:
(i) Find $\small{\big[f(x) \left(A \right) \big]}$. This is the first term.
(ii) Find $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}}$. This is the second term.
(iii) Subtract second term from first term and add the constant of integration.
Let us see a solved example:
Solved Example 23.35
Find $\small{\int{\left[x \cos x \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = x
♦ Let second function be: g(x) = cos x
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\cos x \right]dx}~=~\sin x}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[x \, \sin x \big]}$
• This is the first term.
4. $\small{f'(x)~=~1}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[1\,\left(\sin x \right) \big]dx}~=~-\cos x}$
• This is the second term.
6. So we get:
$\small{\int{\left[x \cos x \right]dx}~=~\text{First term - Second term}~=~x \sin x ~+~\cos x~+~\rm{C}}$
In the above solved example, let us interchange f(x) and g(x). We will write the steps again:
1. Assigning first and second functions:
♦ Let first function be: f(x) = cos x
♦ 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. $\small{\big[f(x) \left(A \right) \big]~=~\big[\cos x \, \left(\frac{x^2}{2} \right) \big]}$
• This is the first term.
4. $\small{f'(x)~=~- \sin x}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[- \sin x\,\left(\frac{x^2}{2} \right) \big]dx}}$
• This is the second term.
6. We see that:
The second term is the integral of yet another product, also with higher power of x. That means, the calculations are becoming lengthy.
• So it is clear that, selection of f(x) and g(x) is important.
The following three rules can be used to select f(x):
(i) One factor is a polynomial function.
•
The other factor is a function other than inverse trigonometric or logarithmic.
•
Then the polynomial function should be taken as f(x).
(ii) One factor is a polynomial function.
•
The other factor is an inverse trigonometric function.
•
Then the inverse trigonometric function should be taken as f(x).
(iii) One factor is a polynomial function.
•
The other factor is a logarithmic function.
•
Then the logarithmic function should be taken as f(x).
In the above solved example, we do integration two times. First in step (2) and later in step (5). But we add the constant of integration only in the final step. Let us see if it makes any difference when the constant is added in every integration step. We will write all the steps again:
1. Assigning first and second functions:
♦ Let first function be: f(x) = x
♦ Let second function be: g(x) = cos x
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\cos x \right]dx}~=~\sin x~+~\rm{C_1}}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[x \,\left( \sin x~+~\rm{C_1}\right) \big]~=~\big[x \, \sin x~+~x\,\rm{C_1} \big]}$
• This is the first term.
4. $\small{f'(x)~=~1}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[1\,\left(\sin x~+~\rm{C_1} \right) \big]dx}~=~-\cos x~+~\rm{C_1}x~+~\rm{C_2}}$
• This is the second term.
6. So we get:
$\small{\int{\left[x \cos x \right]dx}~=~\text{First term - Second term}}$
$\small{~=~\big[x \, \sin x~+~x\,\rm{C_1} \big]~-~\left[-\cos x~+~\rm{C_1}x~+~\rm{C_2} \right]}$
$\small{~=~x \, \sin x~+~x\,\rm{C_1} ~+~\cos x~-~\rm{C_1}x~-~\rm{C_2} }$
$\small{~=~x \, \sin x ~+~\cos x~-~\rm{C_2} }$
$\small{~=~x \, \sin x ~+~\cos x~+~\rm{C} }$
◼ It is clear that, adding constants in the intermediate steps will not make any difference. We need to add a constant in the final step only.
Solved Example 23.36
Find $\small{\int{\left[x e^x \right]dx}}$
Solution:
♦ Let first function be: f(x) = x
♦ Let second function be: g(x) = $\small{e^x}$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[e^x \right]dx}~=~e^x}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[x \,e^x \big]}$
• This is the first term.
4. $\small{f'(x)~=~1}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[1\,\left(e^x \right) \big]dx}~=~e^x}$
• This is the second term.
6. So we get:
$\small{\int{\left[x\, e^x \right]dx}~=~\text{First term - Second term}~=~x \, e^x~-~e^x~+~\rm{C}}$
Solved Example 23.37
Find $\small{\int{\left[x e^{6x} \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = x
♦ Let second function be: g(x) = $\small{e^{6x}}$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[e^{6x} \right]dx}~=~\frac{e^{6x}}{6}}$
(The reader may write all steps involved in the integration process)
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[x \, \left(\frac{e^{6x}}{6} \right) \big]}$
• This is the first term.
4. $\small{f'(x)~=~1}$
5.
$\small{\int{\big[f'(x)\,\left(A \right)
\big]dx}~=~\int{\big[1\,\left(\frac{e^{6x}}{6} \right)
\big]dx}~=~\frac{e^{6x}}{36}}$
(The reader may write all steps involved in the integration process)
• This is the second term.
6. So we get:
$\small{\int{\left[x\,
e^{6x} \right]dx}~=~\text{First term - Second term}~=~x \,
\left(\frac{e^{6x}}{6} \right)~-~\frac{e^{6x}}{36}~+~\rm{C}}$
Solved Example 23.38
Find $\small{\int{\left[\frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = $\small{\sin^{-1}x}$
♦ Let second function be: g(x) = $\small{\frac{x }{\sqrt{1 - x^2}} }$
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. $\small{\big[f(x) \left(A \right) \big]~=~\big[(-1)(\sin^{-1}x) \,\sqrt{1-x^2} \big]}$
• 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:
$\small{\int{\left[\frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \right]dx}~=~\text{First term - Second term}}$
$\small{~=~\big[(-1)(\sin^{-1}x) \,\sqrt{1-x^2} \big]~-~\big[-x \big]~+~\rm{C}}$
$\small{~=~x~-~(\sin^{-1}x) \,\sqrt{1-x^2}~+~\rm{C}}$
Alternate method:
• Put $\small{\sin^{-1}x = \theta}$
• Then we get: $\small{\sin \theta = x}$
• Also, $\small{\frac{d \theta}{dx}~=~\frac{1}{\sqrt{1 -x^2}}}$
$\small{\Rightarrow~\frac{dx}{\sqrt{1 - x^2}}~=~d\theta}$
• So the question becomes: $\small{\int{\left[\sin \theta\,(\theta)\right]dx}}$
Now we can write the usual steps:
1. Assigning first and second functions:
♦ Let first function be: $\small{f(\theta)~=~\theta}$
♦ Let second function be: $\small{g(\theta)~=~\sin \theta}$
2. Finding A:
$\small{A~=~\int{\left[g(\theta) \right]dx}~=~\int{\left[\sin \theta \right]dx}~=~-\cos \theta}$
3. $\small{\big[f(\theta) \left(A \right) \big]~=~\big[(-1)(\theta) \,\cos \theta \big]}$
• This is the first term.
4. $\small{f'(\theta)~=~1}$
5. $\small{\int{\big[f'(\theta)\,\left(A \right) \big]d\theta}~=~\int{\big[1\,\left(-\cos \theta \right) \big]d\theta}~=~-\sin \theta}$
• This is the second term.
6. So we get:
$\small{\int{\left[\sin \theta\,(\theta) \right]dx}~=~\text{First term - Second term}}$
$\small{~=~\sin \theta~-~\theta \cos \theta~+~\rm{C}}$
7. Substituting for $\small{\theta}$, we get:
$\small{\int{\left[\frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \right]dx}~=~x~-~(\sin^{-1} x)\sqrt{1 - x^2}~+~\rm{C}}$
Solved Example 23.39
Find $\small{\int{\left[\log x \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = log x
♦ Let second function be: g(x) = 1
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~ x}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\log x \, (1) \big]~=~\big[\log x \big]}$
• This is the first term.
4. $\small{f'(x)~=~\frac{1}{x}}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[\frac{1}{x}\,\left(x \right) \big]dx}~=~ x}$
• This is the second term.
6. So we get:
$\small{\int{\left[\log x \right]dx}~=~\text{First term - Second term}~=~\log x ~-~ x~+~\rm{C}}$
Solved Example 23.40
Find $\small{\int{\left[e^x \, \sin x \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = $\small{e^x}$
♦ Let second function be: g(x) = $\small{\sin x}$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\sin x \right]dx}~=~-\cos x}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[e^x \, \left(-\cos x\right) \big]}$
• This is the first term.
4. $\small{f'(x)~=~e^x}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[e^x\,\left(-\cos x \right) \big]dx}~=~(-1)\int{\big[e^x\,\left(\cos x \right) \big]dx}}$
• This is similar to the given question. We will do it as a sub group:
*** Beginning of sub group ***
(i) Assigning first and second functions:
♦ Let first function be: f(x) = $\small{e^x}$
♦ Let second function be: g(x) = $\small{\cos x}$
(ii) Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\cos x \right]dx}~=~\sin x}$
(iii) $\small{\big[f(x) \left(A \right) \big]~=~\big[e^x \, \left(\sin x\right) \big]}$
• This is the first term.
(iv) $\small{f'(x)~=~e^x}$
(v) $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[e^x\,\left(\sin x \right) \big]dx}}$
• This is the second term.
(vi) So we get:
$\small{\int{\big[e^x\,\left(\cos x \right) \big]dx}~=~\text{First term - Second term}}$
$\small{~=~\big[e^x \, \left(\sin x\right) \big]~-~\int{\big[e^x\,\left(\sin x \right) \big]dx}~+~\rm{C}}$
$\small{\Rightarrow (-1)\int{\big[e^x\,\left(\cos x \right) \big]dx}~=~\int{\big[e^x\,\left(\sin x \right) \big]dx}~-~\big[e^x \, \left(\sin x\right) \big]~-~\rm{C}}$
*** End of sub group ***
6. So we get:
$\small{\int{\left[e^x \, \sin x \right]dx}~=~\text{First term - Second term}}$
$\small{~=~\big[e^x \, \left(-\cos x\right) \big]~-~\Bigg[\int{\big[e^x\,\left(\sin x \right) \big]dx}~-~\big[e^x \, \left(\sin x\right) \big]~-~\rm{C} \Bigg]}$
$\small{~=~-\big[e^x \, \left(\cos x\right) \big]~-~\int{\big[e^x\,\left(\sin x \right) \big]dx}~+~\big[e^x \, \left(\sin x\right) \big]~+~\rm{C} }$
$\small{\Rightarrow 2\int{\big[e^x\,\left(\sin x \right) \big]dx}~=~-\big[e^x \, \left(\cos x\right) \big]~+~\big[e^x \, \left(\sin x\right) \big]~+~\rm{C} }$
$\small{\Rightarrow \int{\big[e^x\,\left(\sin x \right) \big]dx}~=~\frac{e^x (\sin x \,-\, \cos x)}{2}~+~\rm{C} }$
Solved Example 23.41
Find $\small{\int{\left[x \sin x \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = x
♦ Let second function be: g(x) = $\small{\sin x}$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\sin x \right]dx}~=~-\cos x}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[x \, \left(-\cos x \right) \big]}$
• This is the first term.
4. $\small{~f'(x)~=~1}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[1\,\left(-\cos x \right) \big]dx}~=~-\sin x}$
• This is the second term.
6. So we get:
$\small{\int{\left[x\, \sin x \right]dx}~=~\text{First term - Second term}~=~\big[x \, \left(-\cos x \right) \big]~-~[-\sin x]~+~\rm{C}}$
$\small{~=~-x \cos x ~+~\sin x~+~\rm{C}}$
Solved Example 23.42
Find $\small{\int{\left[x \sin 3x \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = x
♦ Let second function be: g(x) = $\small{\sin 3x}$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\sin 3x \right]dx}~=~-\frac{\cos 3x}{3}}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[x \, \left(-\frac{\cos 3x}{3} \right) \big]~=~ -\frac{x\cos 3x}{3}}$
• This is the first term.
4. $\small{~f'(x)~=~1}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[1\,\left(-\frac{\cos 3x}{3} \right) \big]dx}~=~-\frac{\sin 3x}{9}}$
• This is the second term.
6. So we get:
$\small{\int{\left[x\, \sin 3x \right]dx}~=~\text{First term - Second term}~=~\big[-\frac{x \cos 3x}{3} \big]~-~\left[-\frac{\sin 3x}{9}\right]~+~\rm{C}}$
$\small{~=~-\frac{x\cos 3x}{3} ~+~\frac{\sin 3x}{9}~+~\rm{C}}$
Solved Example 23.43
Find $\small{\int{\left[x^2 e^x \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = $\small{x^2}$
♦ Let second function be: g(x) = $\small{e^x}$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[e^x \right]dx}~=~e^x}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[x^2 \, \left(e^x \right) \big]}$
• This is the first term.
4. $\small{f'(x)~=~2x}$
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[2x\,\left(e^x \right) \big]dx}}$
• This is the second term. We will do it as a sub group:
*** Beginning of sub group ***
(i) Assigning first and second functions:
♦ Let first function be: f(x) = $\small{2x}$
♦ Let second function be: g(x) = $\small{e^x}$
(ii) Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[e^x\right]dx}~=~e^x}$
(iii) $\small{\big[f(x) \left(A \right) \big]~=~\big[2x \, \left(e^x\right) \big]}$
• This is the first term.
(iv) $\small{f'(x)~=~2}$
(v) $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[2\,\left(e^x \right) \big]dx}~=~2 e^x}$
• This is the second term.
(vi) So we get:
$\small{\int{\big[2x\,e^x \big]dx}~=~\text{First term - Second term}}$
$\small{~=~\big[2x \,e^x \big]~-~\big[2e^x \big]}$
*** End of sub group ***
6. So we get:
$\small{\int{\left[x^2\,e^x \right]dx}~=~\text{First term - Second term}}$
$\small{~=~\big[x^2 \, \left(e^x \right) \big]~-~\Bigg[\big[2x \,e^x \big]~-~\big[2e^x \big] \Bigg]~+~\rm{C}}$
$\small{~=~e^x(x^2~-~2x~+~2)~+~\rm{C}}$
Solved Example 23.44
Find $\small{\int{\left[x \, \log x \right]dx}}$
Solution:
1. Assigning first and second functions:
♦ Let first function be: f(x) = log x
♦ Let second function be: g(x) = x
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x\right]dx}~=~\frac{x^2}{2}}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\log x \, \left(\frac{x^2}{2} \right) \big]}$
• 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:
$\small{\int{\left[x \, \log x \right]dx}~=~\text{First term - Second term}~=~\log x \, \left(\frac{x^2}{2} \right)~-~\frac{x^2}{4}~+~\rm{C}}$
In the next section, we will see a few more solved examples.
Previous
Contents
Copyright©2025 Higher secondary mathematics.blogspot.com
No comments:
Post a Comment