Indefinite Integration
If $f$ & $g$ are functions of $x$ such that $g^{\prime}(x)=f(x)$ then,
$$\int f(x) dx = g(x) + c \Leftrightarrow \frac{d}{dx}{g(x) + c} = f(x)$$
where $c$ is called the constant of integration.
Standard Formula:
PYQ-2023-Indefinite_Integration-Q1, PYQ-2023-Definite_Integration-Q1, PYQ-2023-Definite_Integration-Q2, PYQ-2023-Definite_Integration-Q3, PYQ-2023-Definite_Integration-Q4, PYQ-2023-Definite_Integration-Q7, PYQ-2023-Definite_Integration-Q11, PYQ-2023-Definite_Integration-Q12, PYQ-2023-Definite_Integration-Q15, PYQ-2023-Definite_Integration-Q17, PYQ-2023-Definite_Integration-Q18, PYQ-2023-Definite_Integration-Q19, PYQ-2023-Definite_Integration-Q21, PYQ-2023-Definite_Integration-Q22
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$$\int(a x+b)^{n} d x=\frac{(a x+b)^{n+1}}{a(n+1)}+c, n \neq-1$$
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$$\int \frac{\mathrm{dx}}{\mathrm{ax}+\mathrm{b}}=\frac{1}{\mathrm{a}} \operatorname{\ell n}(\mathrm{ax}+\mathrm{b})+\mathrm{c}$$
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$$\int e^{a x+b} d x=\frac{1}{a} e^{a x+b}+c$$
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$$\int \mathrm{a}^{\mathrm{px+q}} \mathrm{dx}=\frac{1}{\mathrm{p}} \frac{\mathrm{a}^{\mathrm{px}+\mathrm{q}}}{\ln \mathrm{a}}+\mathrm{c} ; a>0$$
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$$\int \sin (a x+b) d x=-\frac{1}{a} \cos (a x+b)+c$$
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$$\int \cos (a x+b) d x=\frac{1}{a} \sin (a x+b)+c$$
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$$\quad \int \tan (a x+b) d x=\frac{1}{a} \ln \sec (a x+b)+c$$
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$$\int \cot (a x+b) d x=\frac{1}{a} \ln \sin (a x+b)+c$$
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$$\int \sec ^{2}(a x+b) d x=\frac{1}{a} \tan (a x+b)+c$$
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$$\int \operatorname{cosec}^{2}(a x+b) d x=-\frac{1}{a} \cot (a x+b)+c$$
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$$\int \sec x d x=\ell n (\sec x+\tan x)+c$$
$$ \ell n \tan \left(\frac{\pi}{4}+\frac{x}{2}\right)+c$$
- $$\int \operatorname{cosec} x d x=\ell n(\operatorname{cosec} x-\cot x)+c$$
$$\ell n \tan \frac{x}{2}+c -\ell n(\operatorname{cosec} x+\cot x)+c$$
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$$\quad \int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}+c$$
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$$\int \frac{\mathrm{dx}}{\mathrm{a}^{2}+\mathrm{x}^{2}}=\frac{1}{\mathrm{a}} \tan ^{-1} \frac{\mathrm{x}}{\mathrm{a}}+\mathrm{c}$$
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$$\int \frac{d x}{|x| \sqrt{x^{2}-a^{2}}}=\frac{1}{a} \sec ^{-1} \frac{x}{a}+c$$
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$$\int \frac{\mathrm{dx}}{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}=\operatorname{\ell n}\left[\mathrm{x}+\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}\right]+\mathrm{c}$$
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$$\int \frac{\mathrm{dx}}{\sqrt{\mathrm{x}^{2}-\mathrm{a}^{2}}}=\ell \mathrm{n}\left[\mathrm{x}+\sqrt{\mathrm{x}^{2}-\mathrm{a}^{2}}\right]+\mathrm{c}$$
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$$\quad \int \frac{\mathrm{dx}}{\mathrm{a}^{2}-\mathrm{x}^{2}}=\frac{1}{2 \mathrm{a}} \operatorname{\ell n}\left|\frac{\mathrm{a}+\mathrm{x}}{\mathrm{a}-\mathrm{x}}\right|+\mathrm{c}$$
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$$\int \frac{\mathrm{dx}}{\mathrm{x}^{2}-\mathrm{a}^{2}}=\frac{1}{2 \mathrm{a}} \ell \mathrm{n}\left|\frac{\mathrm{x}-\mathrm{a}}{\mathrm{x}+\mathrm{a}}\right|+\mathrm{c}$$
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$$\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c$$
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$$\int \sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}} \mathrm{dx}=\frac{\mathrm{x}}{2} \sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}+\frac{\mathrm{a}^{2}}{2} \ln \left(\frac{\mathrm{x}+\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}{\mathrm{a}}\right)+\mathrm{c}$$
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$$\int \sqrt{\mathrm{x}^{2}-\mathrm{a}^{2}} \mathrm{dx}=\frac{\mathrm{x}}{2} \sqrt{\mathrm{x}^{2}-\mathrm{a}^{2}}-\frac{\mathrm{a}^{2}}{2} \ln \left(\frac{\mathrm{x}+\sqrt{\mathrm{x}^{2}-\mathrm{a}^{2}}}{\mathrm{a}}\right)+\mathrm{c}$$
Integration by Subsitutions
PYQ-2023-Indefinite_Integration-Q1, PYQ-2023-Indefinite_Integration-Q2, PYQ-2023-Definite_Integration-Q1, PYQ-2023-Definite_Integration-Q4, PYQ-2023-Definite_Integration-Q7, PYQ-2023-Definite_Integration-Q11, PYQ-2023-Definite_Integration-Q13, PYQ-2023-Definite_Integration-Q16, PYQ-2023-Definite_Integration-Q17, PYQ-2023-Definite_Integration-Q22, PYQ-2023-Definite_Integration-Q23
$\quad $ If we subsitute $f(x)=t$, then $$f^{\prime}(x) d x=d t$$
Integration by Part :
PYQ-2023-Definite_Integration-Q8
$$\int(f(x) g(x)) d x=f(x) \int(g(x)) d x-\int (\frac{d}{d x}(f(x)) \int(g(x)) d x) d x$$
Integration of types
PYQ-2023-Indefinite_Integration-Q2
$$ \int \frac{dx}{a x^{2}+b x+c} , \int \frac{dx}{\sqrt{a x^{2}+b x+c}} ,\int \sqrt{a x^{2}+b x+c} d x $$
$\quad $ Make the substitution $$x+\frac{b}{2 a}=t$$
Integration of type
$$ \int \frac{p x+q}{a x^{2}+b x+c} d x, \int \frac{p x+q}{\sqrt{a x^{2}+b x+c}} d x,\int(p x+q) \sqrt{a x^{2}+b x+c} d x $$
$\quad $ Make the substitution $$x+\frac{b}{2 a}=t$$
$\quad $ then split the integral as some of two integrals one containing the linear term and the other containing constant term.
Integration of trigonometric functions
- $$\int \frac{d x}{a+b \sin ^{2} x}$$
$$\int \frac{d x}{a+b \cos ^{2} x}$$
$$\int \frac{d x}{a \sin ^{2} x+b \sin x \cos x+c \cos ^{2} x}$$
$\quad $ put $$\tan x=t$$
- $$\int \frac{d x}{a+b \sin x}$$
$$\int \frac{d x}{a+b \cos x}$$
$$\int \frac{d x}{a+b \sin x+c \cos x}$$
$\quad $ put $$\tan \frac{x}{2}=t$$
- $$\int \frac{a \cdot \cos x+b \cdot \sin x+c}{\ell \cdot \cos x+m \cdot \sin x+n} d x$$
$\quad $ Express $$N r \equiv A(D r)+B \frac{d}{d x}(D r)+c$$
Integration of type:
$$\int \frac{\mathrm{x}^{2} \pm 1}{\mathrm{x}^{4}+K \mathrm{x}^{2}+1} \mathrm{dx}$$
$\quad $ where ${K}$ is any constant.
$\quad $ Divide $Nr$ and $Dr$ by $x^2$ and put $x \mp \frac{1}{x} = t$.
Integration of type:
$$\int \frac{d x}{(a x+b) \sqrt{p x+q}}$$
$$\int \frac{d x}{\left(a x^{2}+b x+c\right) \sqrt{p x+q}} \quad \text { put }\quad p x+q=t^{2}$$
Integration of type:
$$ \int \frac{d x}{(a x+b) \sqrt{p x^{2}+q x+r}} \quad \text { put } \quad x+b=\frac{1}{t} $$
$$ \int \frac{d x}{\left(a x^{2}+b\right) \sqrt{p x^{2}+q}} \quad\text { put }\quad x=\frac{1}{t} $$
Integration of type:
- $$\int \sqrt{\frac{x-\alpha}{\beta-x}} d x $$
$$ \int \sqrt{(x-\alpha)(\beta-x)} \quad\text { put }\quad x=\alpha \cos ^2 \theta+\beta \sin ^2 \theta $$
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$$\int \sqrt{\frac{x-\alpha}{x-\beta}} d x \quad \text { put }\quad x=\alpha \cos ^2 \theta+\beta \sin ^2 \theta $$
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$$ \int \sqrt{(x-\alpha)(x-\beta)} \quad\text { put }\quad x=\alpha \sec ^2 \theta-\beta \tan ^2 \theta$$
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$$\int \frac{d x}{\sqrt{(x-\alpha)(x-\beta)}} \quad \text { put }\quad \mathrm{x}-\alpha=\mathrm{t}^2 \text { or } \mathrm{x}-\beta=\mathrm{t}^2 $$
Indefinite Integration Formula:
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$$ \int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{a}+c$$
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$$\int \frac{d x}{a^2+x^2}=\frac{1}{2} \tan ^{-1} \frac{x}{a}+c$$
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$$\int \frac{d x}{|x| \sqrt{x^2-a^2}}=\frac{1}{a} \sec ^{-1} \frac{x}{a} + c$$
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$$\int \frac{d x}{\sqrt{x^2+a^2}}=\ln \left[x+\sqrt{x^2+a^2}\right]+c$$
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$$\int \frac{d x}{\sqrt{x^2-a^2}}=\ln \left[x+\sqrt{x^2-a^2}\right]+c$$
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$$\int \frac{d x}{a^2-x^2}=\frac{1}{2 a} \ln \left|\frac{a+x}{a-x}\right|+c$$
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$$ \int \frac{d x}{x^2-a^2}=\frac{1}{2 a} \ln \left|\frac{x-a}{x+a}\right|+c$$
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$$\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}+c$$
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$$\int \sqrt{x^2+a^2} d x=\frac{x}{2} \sqrt{x^2+a^2}+\frac{a^2}{2} \ln \left(\frac{x+\sqrt{x^2+a^2}}{a}\right)+c$$
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$$\int \sqrt{x^2-a^2} d x=\frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2} \ln \left(\frac{x+\sqrt{x^2-a^2}}{a}\right)+c$$
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$$\int e^{a x} \cdot \sin b x d x=\frac{e^{a x}}{a^2+b^2}(a \sin b x-b \cos b x)+c$$
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$$\int e^{a x} \cdot \cos b x d x=\frac{e^{a x}}{a^2+b^2}(a \cos b x+b \sin b x)+c$$
Theorems on integration:
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$$ \int c f(x) d x=c \int f(x) d x$$
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$$ \int[f(x) \pm g(x)] d x=\int f(x) d x \pm \int g(x) d x$$
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$$ \int f(x) d x=g(x)+c$$
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$$ \Rightarrow \int f(a x+b) d x=\frac{F(a x+b)}{a}+c, \quad a\ne0 $$
Integration of type $\int \sin m x \cdot \cos n x d x$ :
$\quad$ Case 1. If $m$ and $n$ are even natural numbers then express $\sin m x \cos n x$ in the terms of sines and cosines of multiples of $x$ by using trigonometric results or De’ Moivere’s theorem.
$\quad$ Case 2.
- If $m$ is an odd natural number then put $\cos x=t$.
- If $\mathrm{n}$ is an odd natural number then put $\sin \mathrm{x}=\mathrm{t}$.
- If both $\mathrm{m}$ and $\mathrm{n}$ are odd natural numbers then put either $\sin \mathrm{x}=\mathrm{t}$ or $\cos \mathrm{x}=\mathrm{t}$.
$\quad$ Case 3. When $m+n$ is a negative even integer then put tan $x=t$.
Reduction formula of $\int \tan ^n x d x, \int \cot ^n x d x, \int \sec ^n x d x, \int \operatorname{cosec}^n x d x , n\ne1$:
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If $I_n =\int \tan ^n x d x$, then $$ I_n =\frac{\tan ^{n-1} x}{n-1}-I_{n-2} $$
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If $I_n =\int \cot ^n x d x$, then
$$ I_n=-\frac{\cot ^{n-1} x}{n-1}-I_{n-2} $$
- If $I_n =\int \sec ^n x d x$, then
$$ I_n=\frac{\tan x \sec ^{n-2} x}{n-1}+\frac{n-2}{n-1} I_{n-2} $$
- If $I_n =\int \operatorname{cosec}^n x$, then
$$I_n=\frac{\cot x \operatorname{cosec}^{n-2} x}{n-1}+\frac{n-2}{n-1} I_{n-2}$$