JEE Main Indefinite Integrals Practice Questions With Solutions

Let $\int \frac{2-\tan x}{3+\tan x}\mathrm{d}x=\frac{1}{2}(\alpha x+{\log }_e|\beta \sin x+\gamma \cos x|)+C$\int \frac{2 - tan ⁡ x}{3 + tan ⁡ x} \textrm{ } d x = \frac{1}{2} \left(\right. \alpha x + log_{e} ⁡ \left|\right. \beta sin ⁡ x + \gamma cos ⁡ x \left|\right. \left.\right) + C, where $C$C is the constant of integration. Then $\alpha +\frac{\gamma }{\beta }$\alpha + \frac{\gamma}{\beta} is equal to :

Let $I(x)=\int \frac{6}{{\sin }^{2}x(1-\cot x{)}^2}dx$I \left(\right. x \left.\right) = \int \frac{6}{sin^{2} ⁡ x \left(\right. 1 - cot ⁡ x \left.\right)^{2}} d x. If $I(0)=3$I \left(\right. 0 \left.\right) = 3, then $I\left(\frac{\pi }{12}\right)$I \left(\right. \frac{\pi}{12} \left.\right) is equal to

If $\int \frac{1}{{\mathrm{a}}^2{\sin }^{2}x+{\mathrm{b}}^2{\cos }^{2}x}\mathrm{d}x=\frac{1}{12}{\tan }^{-1}(3\tan x)+$\int \frac{1}{a^{2} sin^{2} ⁡ x + b^{2} cos^{2} ⁡ x} \textrm{ } d x = \frac{1}{12} tan^{- 1} ⁡ \left(\right. 3 tan ⁡ x \left.\right) + constant, then the maximum value of $\mathrm{a}\sin x+\mathrm{b}\cos x$a sin ⁡ x + b cos ⁡ x, is :

If $\int \frac{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}{\sqrt{{\sin }^{3}x{\cos }^{3}x\sin (x-\theta )}}dx=A\sqrt{\cos \theta \tan x-\sin \theta }+B\sqrt{\cos \theta -\sin \theta \cot x}+C$\int \frac{sin^{\frac{3}{2}} ⁡ x + cos^{\frac{3}{2}} ⁡ x}{\sqrt{sin^{3} ⁡ x cos^{3} ⁡ x sin ⁡ \left(\right. x - \theta \left.\right)}} d x = A \sqrt{cos ⁡ \theta tan ⁡ x - sin ⁡ \theta} + B \sqrt{cos ⁡ \theta - sin ⁡ \theta cot ⁡ x} + C, where $C$C is the integration constant, then $AB$A B is equal to

For $x\in (-\frac{\pi }{2},\frac{\pi }{2})$x \in \left(\right. - \frac{\pi}{2} , \frac{\pi}{2} \left.\right), if $y(x)=\int \frac{\mathrm{cosec}x+\sin x}{\mathrm{cosec}x\sec x+\tan x{\sin }^{2}x}dx$y \left(\right. x \left.\right) = \int \frac{cosec x + sin ⁡ x}{cosec x sec ⁡ x + tan ⁡ x sin^{2} ⁡ x} d x, and $\underset{x\to {\left(\frac{\pi }{2}\right)}^{-}}{lim}y(x)=0$\underset{x \rightarrow \left(\left(\right. \frac{\pi}{2} \left.\right)\right)^{-}}{lim} y \left(\right. x \left.\right) = 0 then $y\left(\frac{\pi }{4}\right)$y \left(\right. \frac{\pi}{4} \left.\right) is equal to

$\text{ The integral }\int \frac{(x^8-x^2)\mathrm{d}x}{(x^{12}+3x^6+1){\tan }^{-1}(x^3+\frac{1}{x^3})}\text{ is equal to : }$\textrm{ }\text{The integral}\textrm{ } \int \frac{\left(\right. x^{8} - x^{2} \left.\right) d x}{\left(\right. x^{12} + 3 x^{6} + 1 \left.\right) tan^{- 1} ⁡ \left(\right. x^{3} + \frac{1}{x^{3}} \left.\right)} \textrm{ }\text{is equal to }:\textrm{ }

For $\alpha ,\beta ,\gamma ,\delta \in \mathbb{N}$\alpha , \beta , \gamma , \delta \in \mathbb{N}, if $\int ({\left(\frac{x}{e}\right)}^{2x}+{\left(\frac{e}{x}\right)}^{2x}){\log }_{e}xdx=\frac{1}{\alpha }{\left(\frac{x}{e}\right)}^{\beta x}-\frac{1}{\gamma }{\left(\frac{e}{x}\right)}^{\delta x}+C$\int \left(\right. \left(\left(\right. \frac{x}{e} \left.\right)\right)^{2 x} + \left(\left(\right. \frac{e}{x} \left.\right)\right)^{2 x} \left.\right) log_{e} ⁡ x d x = \frac{1}{\alpha} \left(\left(\right. \frac{x}{e} \left.\right)\right)^{\beta x} - \frac{1}{\gamma} \left(\left(\right. \frac{e}{x} \left.\right)\right)^{\delta x} + C, where $e=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}$e = \sum_{n = 0}^{\infty} \frac{1}{n !} and $\mathrm{C}$C is constant of integration, then $\alpha +2\beta +3\gamma -4\delta$\alpha + 2 \beta + 3 \gamma - 4 \delta is equal to :

If $I(x)=\int e^{{\sin }^{2}x}(\cos x\sin 2x-\sin x)dx$I \left(\right. x \left.\right) = \int e^{sin^{2} x} \left(\right. cos ⁡ x sin ⁡ 2 x - sin ⁡ x \left.\right) d x and $I(0)=1$I \left(\right. 0 \left.\right) = 1, then $I\left(\frac{\pi }{3}\right)$I \left(\right. \frac{\pi}{3} \left.\right) is equal to :

The integral $\int [{\left(\frac{x}{2}\right)}^x+{\left(\frac{2}{x}\right)}^x]\ln \left(\frac{ex}{2}\right)dx$\int \left[\right. \left(\left(\right. \frac{x}{2} \left.\right)\right)^{x} + \left(\left(\right. \frac{2}{x} \left.\right)\right)^{x} \left]\right. ln ⁡ \left(\right. \frac{e x}{2} \left.\right) d x is equal to :

Let $I(x)=\int \frac{(x+1)}{x{(1+x{e}^x)}^2}dx,x>0$I \left(\right. x \left.\right) = \int \frac{\left(\right. x + 1 \left.\right)}{x \left(\left(\right. 1 + x e^{x} \left.\right)\right)^{2}} d x , x > 0. If $\underset{x\to \mathrm{\infty }}{lim}I(x)=0$\underset{x \rightarrow \infty}{lim} I \left(\right. x \left.\right) = 0, then $I(1)$I \left(\right. 1 \left.\right) is equal to :