Antiderivatives by Substitution Method

Integrals can be integrated easily when they are in standard forms. But a given integral may not always be in a standard form. So to reduce it to a standard form, we have to change the variable into a new one by a suitable substitution. Hence the name, Antiderivatives by Substitution Method.

At the time of substitution, we express the given function $x$ in term of a new variable $y$ (say) and replace $dx$ by $\frac{dy}{dx}.dx$.

Form I: Integrals of the forms $\int\{ƒ(x)\}^nƒ'(x)dx$ and $\int\frac{ƒ'(x)}{\{ƒ(x)\}^n}dx$

Here, we substitute $ƒ(x)=y$ so that $ƒ'(x)dx=dy$. Then the given integrals are $\int y^ndy$ and $y^{-n}dy$ which can be easily integrated.

Form II: Integrals of the forms $ƒ(x)g(x)dx$ and $\frac{ƒ(x)}{g(x)}dx$

Sometimes one function is not the derivative of the other, even then we integrate by substituting $g(x)=y$ (say).


Integrate the following:

$\int 3x^2(x^2+1)^3dx$

\[\text{Let }I=\int 3x^2(x^2+1)^3dx\] Put $x^3+1=t$, then, $3x^2dx=dt$ \[\therefore I=\int t^3dt=\frac{t^4}{4}+C\] \[\therefore I=\frac{1}{4}(x^3+1)^4+C\]


$\int\frac{x^{n-1}}{\sqrt{a^n+x^n}}dx$

\[\text{Let }I=\int\frac{x^{n-1}}{\sqrt{a^n+x^n}}dx\] Put $a^n+x^n=t$, then \[nx^{n-1}dx=dt\] \[x^{n-1}dx=\frac{dt}{n}\] \[\therefore I=\frac{1}{n}\int\frac{dt}{\sqrt{t}}\] \[=\frac{1}{n}\int t^{-\frac{1}{2}}dt=\frac{1}{n}\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C\] \[=\frac{2}{n}\sqrt{t}+C\] \[\therefore I=\frac{2}{n}\sqrt{a^n+x^n}+C\]


$\int\frac{(3x+2)dx}{(3x^2+4x+1)^3}$

\[\text{Let }I=\int\frac{(3x+2)dx}{(3x^2+4x+1)^3}\] Put $3x^2+4x+1=t$, then \[(6x+4)dx=dt\] \[(3x+2)dx=\frac{dt}{2}\] \[\therefore I=\frac{1}{2}\int\frac{dt}{t^3}\] \[=\frac{1}{2}\int t^{-3}dt=\frac{1}{2}t^{-2}+C\] \[\therefore I=\frac{1}{2}(3x^2+4x+1)^{-2}+C\]


$\int\frac{(x^2+1)dx}{\sqrt{x^3+3x+4}}$

\[\text{Let }I=\int\frac{(x^2+1)dx}{\sqrt{x^2+3x+4}}\] Put $x^3+3x+4=t$, then, \[(3x^2+3)dx=dt\] \[(x^2+1)dx=\frac{dt}{3}\] \[\therefore I=\frac{1}{3}\int\frac{dt}{\sqrt{t}}=\frac{2}{3}\sqrt{t}+C\] \[\therefore I=\frac{2}{3}\sqrt{x^3+3x+4}+C\]


$\int\frac{1}{x}\sin(\log x)dx$

\[\text{Let }I=\int\frac{1}{x}\sin(\log x)dx\] Put $\log x=t$, then, $\frac{1}{x}dx=dt$. \[\therefore I=\int\sin t dt=-\cos t+C\] \[\therefore I=-\cos(\log x)+C\]


$\int\tan^2\theta\sec^4\theta d\theta$

\[\text{Let }I=\int\tan^2\theta\sec^4\theta d\theta\] \[I=\int\tan^2\theta\sec^2\theta(1+\tan^2\theta)d\theta\] Put $\tan\theta=t$, then, $sec^2\theta d\theta=dt$. \[\therefore I=\int t^2(1+t^2)dt=\int t^2dt+\int t^4dt\] \[=\frac{1}{3}t^3+\frac{1}{5}t^5+C\] \[\therefore I=\frac{1}{3}\tan^3\theta+\frac{1}{5}\tan^5\theta+C\]


$\int\sec x dx$

\[\text{Let }I=\int\sec xdx\] \[I=\int\frac{\sec x(\sec x+\tan x)}{(\sec x+\tan x)}dx\] Put $(\sec x+\tan x)=t$, then, \[(\sec x\tan x+\sec^2x)dx=dt\] \[\sec x(\tan x+\sec x)dx=dt\] \[\therefore I=\int\frac{1}{t}dt=\log t+C\] \[\therefore I=\log(\sec x+\tan x)+C\]


$\int\tan^4xdx$

\[\text{Let }I=\int\tan^4xdx\] \[=\int\tan^2x(\sec^2x-1)dx\] \[=\int\tan^2x\sec^2xdx-\int\tan^2xdx\] \[\int\tan^2x\sec^2xdx-\int(\sec^2x-1)dx\] \[=\int\tan^2x\sec^2xdx-\int\sec^2xdx+\int 1dx\] \[=\int\tan^2x\sec^2x-\tan x+x\] Put $\tan x=t$, then, $\sec^2xdx=dt$. \[\therefore I=\int t^2 dt-\tan x+x\]\[=\frac{1}{3}t^3-\tan x+x+C\] \[\therefore I=\frac{1}{3}\tan^3x-\tan x+x+C\]


$\int\frac{e^x-1}{e^x+1}dx$

\[\text{Let }I=\int\frac{e^x-1}{e^x+1}dx\] \[=\int\frac{\frac{e^x}{e^{\frac{x}{2}}}-\frac{e^x}{e^{\frac{x}{2}}}}{\frac{e^x}{e^{\frac{x}{2}}}+\frac{e^x}{e^{\frac{x}{2}}}}=\int\frac{e^{\frac{x}{2}}-e^{-\frac{x}{2}}}{e^{\frac{x}{2}}+e^{-\frac{x}{2}}}\] Put $e^{\frac{x}{2}}+e^{-\frac{x}{2}}=t$, then, \[\left(e^{\frac{x}{2}}×\frac{1}{2}-e^{-\frac{x}{2}}×\frac{1}{2}\right)dx=dt\] \[=\left(e^{\frac{x}{2}}-e^{-\frac{x}{2}}\right)dx=2dt\] \[\therefore I=\int\frac{2dt}{t}=2\log t+C\] \[\therefore I=2\log(e^{\frac{x}{2}}-e^{-\frac{x}{2}})+C\]


Trigonometrical Substitution

In this section, we shall consider integrals which involve $a^2-x^2$, $a^2+x^2$ or $x^2-a^2$. Substituting $x=a\sin\theta$ turns $a^2-x^2$ into $a^2\cos^2\theta$; $x=a\tan\theta$ turns $a^2+x^2$ into $a^2\sec^2\theta$ and $x=a\sec\theta$ turns $x^2-a^2$ into $a^2\tan^2\theta$. These substitutions will make the resulting function easily integrable.

But if the integral involves $\frac{1}{\sqrt{x^2+a^2}}$ or $\frac{1}{\sqrt{x^2-a^2}}$ only, we can also follow the following method besides the trigonometrical substitution. \[\text{Let }x+\sqrt{x^2+a^2}=y\] \[\text{so that }\left(1+\frac{1}{2\sqrt{x^2+a^2}}2x\right)dx=dy\] \[\frac{x+\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}}dx=dy\] \[\frac{dx}{\sqrt{x^2+a^2}}=\frac{dy}{y}\] Integrating both sides will give the required result. Similarly, for second, we put $x+\sqrt{x^2-a^2}=y$ and the result will be obtained in the same way.


Find the indefinite integrals:

$\int\frac{x^2dx}{\sqrt{a^2-x^2}}$

\[\text{Let }I=\int\frac{x^2dx}{\sqrt{a^2-x^2}}\] Put $x=a\sin\theta$, then, $dx=a\cos\theta d\theta$. \[\therefore I=\int\frac{a^2\sin^2\theta.a\cos\theta d\theta}{\sqrt{a^2-a^2\sin^2\theta}}\] \[=\int\frac{a^3\sin^2\theta\cos\theta d\theta}{a\cos\theta}\] \[=a^2\int\sin^2\theta d\theta=\frac{a^2}{2}\int 2\sin^2\theta d\theta\] \[=\frac{a^2}{2}\int(1-\cos 2\theta)d\theta\] \[=\frac{a^2}{2}\int 1d\theta-\frac{a^2}{2}\int \cos 2\theta d\theta\] \[=\frac{a^2}{2}\theta-\frac{a^2}{2}\frac{\sin 2\theta}{2}+C\] \[=\frac{a^2}{2}\theta-\frac{a^2}{2}\sin\theta\cos\theta+C\] Here, \[\sin\theta=\frac{x}{a}=\frac{p}{h}\] \[\therefore b=\sqrt{h^2-p^2}=\sqrt{a^2-x^2}\] \[\therefore \cos\theta=\frac{b}{h}=\frac{\sqrt{a^2-x^2}}{a}+C\] \[\therefore I=\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)-\frac{a^2}{2}\frac{x}{a}\frac{\sqrt{a^2-x^2}}{a}+C\] \[=\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)-\frac{x}{2}\sqrt{a^2-x^2}+C\]


$\int\frac{dx}{(x^2-a^2)^{\frac{3}{2}}}$

\[\text{Let }I=\int\frac{dx}{(x^2-a^2)^{\frac{3}{2}}}\] Put $x=a\sec\theta$, then, $dx=a\sec\theta\tan\theta d\theta$. \[\therefore I=\int\frac{a\sec\theta\tan\theta d\theta}{(a^2\sec^2\theta-a^2)^{\frac{3}{2}}}\] \[=\int\frac{a\sec\theta\tan\theta d\theta}{a^3\tan^3\theta}=\frac{1}{a^2}\int\frac{\sec\theta}{\tan^2\theta}d\theta\] \[=\frac{1}{a^2}\int \frac{\cos\theta}{\sin^2\theta}d\theta\] Put $\sin\theta=t$, then, $\cos\theta=dt$ \[\therefore I=\frac{1}{a^2}\int\frac{dt}{t^2}=-\frac{1}{a^2t}+C\] \[\therefore I=-\frac{1}{a^2\sin\theta}+C\]\[=-\frac{1}{a^2}\operatorname{cosec}\theta+C\] Here, \[\sec\theta=\frac{x}{a}=\frac{h}{b}\] \[\therefore p=\sqrt{h^2-b^2}=\sqrt{x^2-a^2}\] \[\therefore \operatorname{cosec}\theta=\frac{x}{\sqrt{x^2-a^2}}\] \[\therefore I=-\frac{x}{a^2\sqrt{x^2-a^2}}+C\]


$\int\frac{dx}{x\sqrt{x^2+1}}$

\[\text{Let }I=\int\frac{dx}{x\sqrt{x^2+1}}\] Put $x=\tan\theta$, then, $dx=\sec^2\theta d\theta$. \[\therefore I=\int\frac{\sec^2\theta d\theta}{\tan\theta\sqrt{\tan^2\theta+1}}=\int\frac{\sec^2 \theta d\theta}{\tan\theta\sec\theta}\] \[=\int\frac{\sec\theta}{\tan\theta}d\theta= \int \operatorname{cosec} \theta d\theta\] \[=\log(\operatorname{cosec}\theta-\cot\theta)+C\] Here, \[\tan\theta=\frac{x}{1}=\frac{p}{b}\] \[\therefore h=\sqrt{p^2+b^2}=\sqrt{x^2+1}\] \[\therefore \operatorname{cosec}\theta=\frac{\sqrt{x^2+1}}{x} \text{ and }\cot\theta=\frac{1}{x}\] \[\therefore I=\log\left(\frac{\sqrt{x^2+1}}{x}-\frac{1}{x}\right)+C\] \[=\log\left(\frac{\sqrt{x^2+1}-1}{x}\right)+C\]


$\int\frac{\tan^{-1}x}{1+x^2}dx$

\[\text{Let }I=\int\frac{\tan^{-1}x}{1+x^2}dx\] Put $\tan^{-1}x=t$, then, \[\frac{1}{1+x^2}dx=dt\] \[\therefore I=\int t dt=\frac{t^2}{2}+C\] \[\therefore I=\frac{1}{2}(\tan^{-1}x)^2+C\]


$\int\sqrt{\frac{x}{a-x}}dx$

\[\text{Let }I=\int\sqrt{\frac{x}{a-x}}dx\] Put $x=a\sin^2\theta$, then, $dx=2a\sin\theta\cos\theta d\theta$. \[\therefore I=\int\sqrt{\frac{a\sin^2\theta}{a-a\sin^2\theta}}2a\sin\theta\cos\theta d\theta\] \[=\int\frac{\sin\theta}{\cos\theta}2a\sin\theta\cos\theta d\theta\] \[=a\int 2\sin^2\theta d\theta\] \[=a\int (1-\cos 2\theta)d\theta\] \[=a\int 1d\theta-a\int\cos 2\theta d\theta\] \[=a\theta-\frac{a\sin 2\theta}{2}+C\] \[=a\theta-a\sin\theta\cos\theta+C\] Here, \[\sin\theta=\sqrt{\frac{x}{a}}=\frac{p}{h}\] \[\therefore b=\sqrt{h^2-p^2}=\sqrt{a-x}\] \[\therefore \cos\theta=\sqrt{\frac{a-x}{a}}\] \[\therefore I=a\sin^{-1}\sqrt{\frac{x}{a}}-a\sqrt{\frac{x}{a}}.\sqrt{\frac{a-x}{a}}+C\] \[=a\sin^{-1}\sqrt{\frac{x}{a}}-\sqrt{x(a-x)}+C\]

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