Integration by Substitution/Replacement
The integration by substitution rule applies if the integrand consists two parts (functions) in which one function is derivative of other (in some form). The two general structure of this form is
\( \int f(x)^n.f(x)' dx\) or \( \int \frac{f(x)' dx}{f(x)^n} \)
That is, if a one part is the derivative of another via the chain rule, we let u to denote a likely candidate for the inner part as f(x), then translate the given function so that it is written entirely in terms of u, with no x remaining in the expression.
If we can integrate this new function of u, then the anti-derivative of the original function is obtained by replacing u by the equivalent expression in x.
It can be further classified into three sub categories
- General substitution
- Powers of sine and cosine
- Trigonometric Substitutions
- \( \int \frac{dx}{\text{Quadratic}} \)
- \( \int \frac{\text{linear}}{\text{Quadratic}} dx \)
- \( \int \frac{dx} {\sqrt{\text{Quadratic}}} dx \)
- \( \int \frac{\text{linear} }{\sqrt{\text{Quadratic}}} dx \)
- \( \int \frac{dx}{x^2-a^2} dx =\frac{1}{2a} \log \left | \frac{x-a}{x+a}\right |+c \)
- \( \int \frac{dx}{a^2-x^2} dx =\frac{1}{2a} \log \left | \frac{a+x}{a-x}\right |+c \)
- \( \int \frac{dx}{x^2+a^2} dx =\frac{1}{a} \tan^{-1} \left ( \frac{x}{a}\right )+c \)
- \( \int \frac{dx}{\sqrt{a^2-x^2}} dx =\sin^{-1} \left ( \frac{x}{a}\right )+c \)
- \( \int \frac{dx}{\sqrt{x^2-a^2}} dx =\log \left | x+ \sqrt{x^2-a^2} \right |+c \)
- \( \int \frac{dx}{\sqrt{x^2+a^2}} dx =\log \left | x+ \sqrt{x^2+a^2} \right |+c \)
- \( \int \frac{dx}{x\sqrt{x^2-a^2}} dx =\frac{1}{a} \sec^{-1} \left ( \frac{x}{a}\right )+c \)
Solved Examples: General substitution
Evaluate Following Integrals
- \( \int (x^2-5)^7 2x dx\)
Solution 👉 Click Here
Solution
Here,
derivative of x2 is 2x, therefore, we let
\( u=x^2\)
Now
\( \frac{du}{dx}= 2x\) or du=2xdx
Since we have exactly 2xdx in the original integral, we can replace it by du, hence, we have
\( \int (x^2-5)^7 2xdx\)
or
\( \int (u-5)^7 du\)
or
\( \int (u-5)^7 du\)
or
\( \frac{(u-5)^8}{8}+c\)
or
\( \frac{(x^2-5)^8}{8}+c\)
- \(\int 2x (\cos x^2) \)dx
Solution 👉 Click Here
Solution
Here,
Derivative of \(x^2\) is 2x, therefore, we let
\( u=x^2\)
Now
\(\frac{du}{dx}= 2x\) or du=2xdx
Since we have exactly 2xdx in the original integral, we can replace it by du, hence, we have
\(\int 2x (\cos x^2) dx\)
or
\(\int (\cos u) du \)
or
\(\int 2x (\cos x^2) dx=\sin u+c\)
or
\(\int 2x (\cos x^2) dx=\sin (x^2)+c\)
- \(\int (ax+b)^ndx\), assuming that a and b are constants, \( a\ne 0\), and n is a positive integer.
Solution 👉 Click Here
Solution
We let u=ax+b so du=adx or dx=\(\frac{du}{a}\). Then
\(\int (ax+b)^ndx\)
or
\(\int u^n \frac{du}{a}\)
or
\(\frac{1}{a} \int u^n du\)
or
\(\frac{1}{a} \frac{u^{n+1}}{n+1}+c\)
or
\(\frac{1}{a(n+1)}(ax+b)^{n+1}+c\)
- \( \int \sin(ax+b)dx\), assuming that a and b are constants and \(a\ne 0\).
Solution 👉 Click Here
Solution
We let
u=ax+b so du=adx or dx=\( \frac{du}{a}\)
Then
\( \int \sin(ax+b)dx\)
or
\( \int \sin(u) \frac{du}{a}\)
or
\( \frac{1}{a} (-\cos u)+c\)
or
\( -\frac{1}{a} \cos (ax+b)+c \)
Evaluate the following Intrgral
| SN |
Question |
Answer |
| 1 |
\( \int \sin^2 x dx\) |
\( \frac{x}{2}-\frac{\sin(2x)}{4}+c\) |
| 2 |
\( \int \sin^3 x dx\) |
\( -\cos x+\frac{\cos^3 x}{3}+c\) |
| 3 |
\( \int \sin^4 x dx\) |
\(\frac{3x}{8}-\frac{\sin(2x)}{4}+\frac{\sin(4x)}{32}+c\) |
| 4 |
\( \int \cos^2 x \sin^3 x dx\) |
\(\frac{\cos^5 x}{5}-\frac{\cos^3 x}{3}+c\) |
| 5 |
\( \int \cos^3 x dx\) |
\(-\sin x-\frac{\sin^3 x}{3}+c\) |
| 6 |
\( \int \cos^3 x \sin^2 xdx\) |
\(\frac{\sin^3 x}{3}-\frac{\sin^5 x}{5}+c\) |
Solved Examples: Powers of sine and cosine
Evaluate the following integrals
- \( \int \sin^5xdx\).
Solution 👉 Click Here
Solution
Here, integrand consists products of the sine, sine with odd exponent, so it can be integrated by using substitution of trigonometric identities. Thus,
\( \int \sin^5xdx\)
or
\( \int \sin x \sin^4xdx\)
or
\( \int \sin x (1- \cos^2x)^2 dx\)
Now, we use
\( u=\cos x\) then \(du=-\sin x dx\)
Hence, we have
\( \int \sin^5xdx\)
or
\(\int (1-u^2)^2 (-du)\)
or
\(-\int (1-2u^2+u^4) du\)
or
\(\int \sin^5xdx=-\int (1-2u^2+u^4) du\)
or
\(-\left (u-\frac{2u^3}{3}+\frac{u^5}{5}+c \right )\)
or
\(-\left (\cos x-\frac{2\cos^3 x}{3}+\frac{\cos^5x}{5}+c \right )\)
- \( \int \sin^2x \cos^2x dx\).
Solution 👉 Click Here
Solution
Here,integrand consists products of the sine and cosine, sine with even exponent, so it can be integrated by using substitution of trigonometric identities. Thus,
\(\int \sin^2x \cos^2 dx\)
or
\( \int \frac{1}{8} (1-\cos 4x) dx \)
or
\( \frac{1}{8}\int (1-\cos 4x) dx\)
or
\( \frac{1}{8}\left (x-\frac{\sin 4x}{4} \right )+c \)
- \( \int \frac{\tan ^4 \sqrt{x} \sec ^2 \sqrt{x}}{\sqrt{x} } dx \)
Solution 👉 Click Here
Solution
यस प्रश्नमा \(\tan \sqrt{x}=u \) मानौ , then \(\frac{\sec ^2 \sqrt{x} }{ \sqrt{x}} dx= 2du\)
Now, the integral is
\( \int \frac{\tan ^4 \sqrt{x} \sec ^2 \sqrt{x}}{\sqrt{x} } dx \)
or\( \int (\tan ^4 \sqrt{x}) \left ( \frac{ \sec ^2 \sqrt{x}}{ \sqrt{x}} dx \right ) \)
or\(\int (u^4) (2 du)\)
or\(\frac{2}{5} u^5+c\)
or\(\frac{2}{5} (\tan \sqrt{x})^5+c\)
or\(\frac{2}{5} \tan^5 \sqrt{x}+c\)
- \( \int \csc^6 dx \)
Solution 👉 Click Here
Solution
यस प्रश्नमा \(\cot x=u \) मानौ ,then \( \csc^2x dx=-du\)
Now, the integral is
\( \int \csc^6 dx \)
or\( \int \csc^4x \csc^2 x dx \)
or\( \int (\csc^2x)^2 \csc^2 x dx \)
or\( \int (1+\cot^2x)^2 \csc^2 x dx \)
or\( \int (1+u^2)^2 (-du) \)
or\( -\int (1+2u^2+u^4) du \)
or\( -\left ( u+\frac{2}{3}u^3+\frac{1}{5}u^5+c \right) \)
or\( -\left ( u+\frac{2}{3}\cot^3 x+\frac{1}{5}\cot^5x \right) +c\)
- \( \int \frac{\sin(x+a)}{\sin(x-a)} dx \)
Solution 👉 Click Here
Solution
यस प्रश्नमा \(x-a=u \) मानौ , then \( dx=du\)
Now, the integral is
\( \int \frac{\sin(x+a)}{\sin(x-a)} dx \)
or\( \int \frac{\sin(a+u+a)}{\sin(u)} du \)
or\( \int \frac{\sin(u+2a)}{\sin(u)} du \)
or\( \int \frac{\sin(u) \cos(2a)+\cos(u) \sin(2a)}{\sin(u)} du \)
or\( \int (\cos(2a)+\cot(u) \sin(2a) )du \)
or\( u \cos(2a)+ \log (\sin(u)) \sin(2a) +c \)
or\( (x-a) \cos(2a)+ \log (\sin(x-a)) \sin(2a) +c \)
Trigonometric Substitutions
This type of substitution is usually indicated when the function of integrate contains a polynomial expression to use the fundamental trigonometric identity as
\(\cos^2x+\sin^2x =1\) or \(\cos^2 x=1-\sin^2 x \)
\(\sec^2x-\tan^2x=1 \) or \( \sec^2x=1+\tan^2x \)
\(\sec^2x-\tan^2x=1 \) or \( \tan^2x=\sec^2x-1\)
Therefore
- if it contains \( 1-x^2\), we substitute \(x=\sin u\)
- if it contains \(1+x^2\) we substitute \(x=\tan u\) and
- if it contains \(x^2-1\), we substitute \(x=\sec u\).
Sometimes, we need something a bit different to handle constants other than one. Thus, key points are as follows
- Put \( x=a\sin u \) for integration involving \( a^2-x^2\)
- Put \( x=a\tan u \) for integration involving \( a^2+x^2\)
- Put \( x=a\sec u \) for integration involving \( x^2-a^2 \)
Solved Examples: Trigonometric Substitutions
Evaluate following integrals
- \( \int \sqrt{1-x^2} dx\)
Solution 👉 Click Here
Solution
This question is in the form \(1-x^2\), so we substitute, \(x=\sin u\), then \( dx=\cos u du\), thus we have
\(\int \sqrt{1-x^2} dx\)
or
\( \int \sqrt{1-\sin^2 u} \cos u du \)
or
\( \int \cos^2 u du\)
or
\( \frac{1+\cos 2 u}{2} du\)
or
\( \frac{u}{2}+\frac{\sin 2 u}{4} +c \)(A)
Here
\(x=\sin u \)
or
\( u=\sin^{-1}x \)(1)
Next
\(\sin 2u =2 \sin u \cos u \)
or
\( \sin 2u=2 \sin \sin^{-1}x \sqrt{1-\sin ^2 u} \)
or
\( \sin 2u=2 x \sqrt{1-x^2} \)(2)
Thus, using (1) and (2),the solution (A) is
\( \frac{u}{2}+\frac{\sin 2 u}{4} +c \)
or
\( \frac{\sin^{-1} x}{2}+\frac{2x \sqrt{1-x^2}}{4} +c \)
- \( \int \sqrt{4-9x^2} dx \)
Solution 👉 Click Here
The question is in the form of \(a^2-x^2\), so we let \(x= a \sin u\)
\(3x= 2 \sin u\)
Then
\(x= \frac{2}{3} \sin u\)
or\(dx= \frac{2}{3} \cos u du\)
Hence the integral is
\( \int \sqrt{4-9x^2} dx \)
or\( \int \sqrt{4-(3x)^2} dx \)
or\( \int \sqrt{(2^2)-(2 \sin u)^2} ( \frac{2}{3} \cos u du) \)
or\( \int \sqrt{2^2(1-\sin^2 u)} ( \frac{2}{3} \cos u du) \)
or\( \int (2 \cos u) ( \frac{2}{3} \cos u du) \)
or\( \int \frac{4}{3} \cos^2 u du \)
or\( \frac{4}{3} \int \frac{1+ \cos 2 u}{2} du \)
or\( \frac{2}{3} \int (1+ \cos 2 u) du \)
or\( \frac{2}{3} \left [u+ \frac{\sin 2 u}{2} \right ] +c \)
or\( \frac{2}{3} \left [ \frac{2u+\sin 2 u}{2} \right ] +c \)
or\( \frac{1}{3} (2u+\sin 2 u) +c \)(A)
Here
\(x=\frac{2}{3} \sin u \)
or
\( u=\sin^{-1}(\frac{3x}{2}) \)(1)
Next
\(\sin 2u =2 \sin u \cos u \)
or
\( \sin 2u=2 \sin \sin^{-1}(\frac{3x}{2}) \sqrt{1-\sin ^2 u} \)
or
\( \sin 2u=2 (\frac{3x}{2}) \sqrt{1-(3x/2)^2} \)
or
\( \sin 2u=3x \frac{1}{2} \sqrt{4-9x^2} \)
or
\( \sin 2u= \frac{3x}{2} \sqrt{4-9x^2} \)(2)
Thus, using (1) and (2),the solution (A) is
\( \frac{1}{3} (2u+\sin 2 u) +c \)
or \( \frac{1}{3} \left ( 2 \sin^{-1}(\frac{3x}{2}) +\frac{3x}{2} \sqrt{4-9x^2} \right ) +c \)
or \( \frac{2}{3} \sin^{-1}(\frac{3x}{2})+ \frac{x}{2} \sqrt{4-9x^2} +c \)
- \( \int \frac{x^2}{\sqrt{4-x^2}} dx\)
Solution 👉 Click Here
Solution
The question is in the form of \(a^2-x^2\), so we let \(x= a \sin u\)
\(x= 2 \sin u\)
Then
\(dx= 2 \cos u du\)
Hence the integral is
\( \int \frac{x^2}{\sqrt{4-x^2}}dx\)
or\( \int \frac{4 \sin ^2 u}{\sqrt{4-4 \sin ^2 u}} (2 \cos u du)\)
or\( \int \frac{4 \sin ^2 u}{2 \sqrt{1- \sin ^2 u}} (2 \cos u du)\)
or\( \int \frac{4 \sin ^2 u}{2 \cos u} (2 \cos u du)\)
or\( \int (4 \sin ^2 u) du \)
or\( 4 \int \frac{1- \cos 2 u}{2} du \)
or\( 2 \int (1- \cos 2 u) du \)
or\( 2 \left [u- \frac{\sin 2 u}{2} \right ] +c \)
or\( 2 \left [ \frac{2u-\sin 2 u}{2} \right ] +c \)
or\( (2u-\sin 2 u) +c \)(A)
Here
\(x=2 \sin u \)
or
\( u=\sin^{-1}(\frac{x}{2}) \)(1)
Next
\(\sin 2u =2 \sin u \cos u \)
or
\( \sin 2u=2 \sin \sin^{-1}(\frac{x}{2}) \sqrt{1-\sin ^2 u} \)
or
\( \sin 2u=2 (\frac{x}{2}) \sqrt{1-(x/2)^2} \)
or
\( \sin 2u=x \frac{1}{2} \sqrt{4-x^2} \)
or
\( \sin 2u= \frac{x}{2} \sqrt{4-x^2} \)(2)
Thus, using (1) and (2),the solution (A) is
\( (2u-\sin 2 u) +c \)
or \( \left ( 2 \sin^{-1}(\frac{x}{2}) -\frac{x}{2} \sqrt{4-x^2} \right ) +c \)
or \( 2 \sin^{-1}(\frac{x}{2})- \frac{x}{2} \sqrt{4-x^2} +c \)
- \( \int \frac{\sqrt{x}}{\sqrt{a^3-x^3}} dx \)
Solution 👉 Click Here
Solution
यस प्रश्नमा
\(x^{3/2}=u \) मानौ , then \( \sqrt{x} dx=\frac{2}{3}du\)
Now, the integral is
\( \int \frac{\sqrt{x}}{\sqrt{a^3-x^3}} dx \)
or \( \int \frac{\sqrt{x}}{\sqrt{(a^{3/2})^2-(x^{3/2})^2}} dx \)
or \( \int \frac{ \frac{2}{3} du}{\sqrt{(a^{3/2})^2-(u)^2}} \)
or \( \frac{2}{3} \int \frac{ du}{\sqrt{(a^{3/2})^2-(u)^2}} \)
Now, \( \int \frac{dx}{\sqrt{a^2-x^2}} dx =\sin^{-1} \left ( \frac{x}{a}\right )+c \) सुत्रको प्रयोग गरौ ।
Then
\( \frac{2}{3} \int \frac{ du}{\sqrt{(a^{3/2})^2-(u)^2}} \)
or \( \frac{2}{3} \sin^{-1} \left ( \frac{u}{a^{3/2}}\right )+c \)
or \( \frac{2}{3} \sin^{-1} \left ( \frac{x^{3/2}}{a^{3/2}}\right )+c \)
Evaluate the following Intrgral
| SN |
Question |
Answer |
| 1 |
\( \int{\frac{1}{\sqrt{x^2-a^2}}}dx\) |
\( \int{\frac{x}{\sqrt{9+x^2}}}dx\) |
| 2 |
\( \int \sqrt{x^2-1} dx\) |
\( \frac{x\sqrt{x^2-1}}{2}-\frac{ln |x+\sqrt{x^2-1}|}{2}+c\) |
| 3 |
\( \int \sqrt{9+4x^2} dx\) |
\( \frac{x\sqrt{9+4x^2}}{2}+ \frac{9}{4} \ln |2x+\sqrt{9+4x^2}|+c\) |
| 4 |
\( \int x \sqrt{1-x^2} dx\) |
\( -\frac{(1-x^2)^{2/3}}{3}+ c\) |
| 5 |
\( \int x^2 \sqrt{1-x^2} dx\) |
\( \frac{\sin^{-1}}{8}-\frac{sin(4 \sin^1{-1}x)}{32}+ c\) |
| 6 |
\(\int \frac{1}{\sqrt{1+x^2}} dx\) |
\(\ln |x+\sqrt{1+x^2}| + c \) |
| 7 |
\(\int \frac{x^2}{\sqrt{4-x^2}} dx\) |
\(2 \sin^{-1} (x/2) -\frac{x\sqrt{4-x^2}}{2} + c\) |
| 8 |
\(\int \frac{x^3}{\sqrt{4x^2-1}} dx\) |
\(\frac{(2x^2+1) \sqrt{4x^2-1}}{24}+ c\) |
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