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JEE MAIN Previous Year Questions (PYQs)

JEE MAIN Indefinite Integration PYQ

JEE MAIN PYQ
Let $\mathrm{I}(x)=\int \frac{d x}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}}$. If $\mathrm{I}(37)-\mathrm{I}(24)=\frac{1}{4}\left(\frac{1}{\mathrm{~b}^{\frac{1}{13}}}-\frac{1}{\mathrm{c}^{\frac{1}{13}}}\right), \mathrm{b}, \mathrm{c} \in \mathcal{N}$, then $3(\mathrm{~b}+\mathrm{c})$ is equal to





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (23 January Morning Shift) PYQ

Solution

Let $t=\frac{x-11}{x+15}$

Then, $dx=\frac{26}{(1-t)^2}dt$

Also, $(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}=t^{\frac{11}{13}}(x+15)^2$

and $x+15=\frac{26}{1-t}$

So, $I=\int \frac{\frac{26}{(1-t)^2}dt}{t^{\frac{11}{13}}\cdot \frac{26^2}{(1-t)^2}}$

$I=\frac{1}{26}\int t^{-\frac{11}{13}}dt$

$I=\frac{1}{26}\cdot\frac{13}{2}t^{\frac{2}{13}}$

$I=\frac14\left(\frac{x-11}{x+15}\right)^{\frac{2}{13}}$

Now, $I(37)-I(24)=\frac14\left[\left(\frac{26}{52}\right)^{\frac{2}{13}}-\left(\frac{13}{39}\right)^{\frac{2}{13}}\right]$

$=\frac14\left[\left(\frac12\right)^{\frac{2}{13}}-\left(\frac13\right)^{\frac{2}{13}}\right]$

$=\frac14\left[\frac{1}{4^{\frac{1}{13}}}-\frac{1}{9^{\frac{1}{13}}}\right]$

So, $b=4,\ c=9$

$3(b+c)=3(4+9)=39$

$\boxed{39}$
JEE MAIN PYQ
The integral $\displaystyle \int (1+x-\frac{1}{x})e^{x+\frac{1}{x}}\,dx$ is equal to





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2014 (Offline) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle \int x^{5}\,e^{-4x^{3}}\,dx=\dfrac{1}{48}\,e^{-4x^{3}}\,f(x)+C$, where $C$ is a constant of integration, then $f(x)$ is equal to –





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (10 January Evening Shift) PYQ

Solution

JEE MAIN PYQ
Let $\int x^3 \sin x \mathrm{~d} x=g(x)+C$, where $C$ is the constant of integration. If $8\left(g\left(\frac{\pi}{2}\right)+g^{\prime}\left(\frac{\pi}{2}\right)\right)=\alpha \pi^3+\beta \pi^2+\gamma, \alpha, \beta, \gamma \in Z$, then $\alpha+\beta-\gamma$ equals :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (23 January Evening Shift) PYQ

Solution

JEE MAIN PYQ
If $\int f(x)\,dx=\psi(x)$, then $\int x^{5}f(x^{3})\,dx$ is equal to





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2013 (Offline) PYQ

Solution

JEE MAIN PYQ
Let $I(x)=\displaystyle \int \frac{x^{2}\big(x\sec^{2}x+\tan x\big)}{(x\tan x+1)^{2}}\,dx.$ If $I(0)=0$, then $I\!\left(\frac{\pi}{4}\right)$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (6 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle \int \frac{\sqrt{\,1-x^{2}\,}}{x^{4}}\,dx = A(x)\left(\sqrt{\,1-x^{2}\,}\right)^{m} + C$, for a suitable chosen integer $m$ and a function $A(x)$, where $C$ is a constant of integration, then $(A(x))^{m}$ equals :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (11 January Morning Shift) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle \int (\sin x)^{-\frac{11}{2}} (\cos x)^{\frac{5}{2}} dx = -\frac{p_1}{q_1}(\cot x)^{\frac{9}{2}} - \frac{p_2}{q_2}(\cot x)^{\frac{5}{2}} - \frac{p_3}{q_3}(\cot x)^{\frac{1}{2}} + \frac{p_4}{q_4}(\cot x)^{-\frac{3}{2}} + C$, where $p_i, q_i$ are positive integers with $\gcd(p_i,q_i)=1$ for $i = 1,2,3,4$ and $C$ is the constant of integration, then $\displaystyle \frac{15p_1p_2p_3p_4}{q_1q_2q_3q_4}$ is equal to ______.





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

$\displaystyle \int (\tan x)^{-\frac{11}{2}} \sec^8 x , dx$ $= \int (\tan x)^{-\frac{11}{2}} (1+\tan^2 x)^3 \sec^2 x , dx$ Put $\tan x = t$ $\Rightarrow \int t^{-\frac{11}{2}} (1+t^2)^3 , dt$ $= \int \left(t^{-\frac{11}{2}} + 3t^{-\frac{7}{2}} + 3t^{-\frac{3}{2}} + t^{\frac{1}{2}}\right) dt$ $= -\frac{2}{9}t^{-\frac{9}{2}} - \frac{6}{5}t^{-\frac{5}{2}} - 2t^{-\frac{1}{2}} + \frac{2}{3}t^{\frac{3}{2}} + C$ Back substitute $t = \cot x$ $\Rightarrow p_1 = 2,; p_2 = 6,; p_3 = 2,; p_4 = 2$ $q_1 = 9,; q_2 = 5,; q_3 = 1,; q_4 = 3$ $\displaystyle \frac{15p_1p_2p_3p_4}{q_1q_2q_3q_4} = \frac{15 \cdot 2 \cdot 6 \cdot 2 \cdot 2}{9 \cdot 5 \cdot 1 \cdot 3} = 16$
JEE MAIN PYQ
If $f\left( {{{x - 4} \over {x + 2}}} \right) = 2x + 1,$ (x $ \in $ R $-${1, $-$ 2}), then $\int f \left( x \right)dx$ is equal to :
(where C is a constant of integration)





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2018 (15 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle \int \frac{x+1}{\sqrt{2x-1}}\,dx = f(x)\,\sqrt{2x-1}+C$, where $C$ is a constant of integration, then $f(x)$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (11 January Evening Shift) PYQ

Solution

JEE MAIN PYQ
The integral $\int {{{{e^{3{{\log }_e}2x}} + 5{e^{2{{\log }_e}2x}}} \over {{e^{4{{\log }_e}x}} + 5{e^{3{{\log }_e}x}} - 7{e^{2{{\log }_e}x}}}}} dx$, x > 0, is equal to : (where c is a constant of integration)





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (25 February Evening Shift) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle \int \frac{2x+5}{\sqrt{7-6x-x^{2}}},dx = A\sqrt{7-6x-x^{2}} + B\sin^{-1}!\left(\frac{x+3}{4}\right) + C$ (where $C$ is a constant of integration), then the ordered pair $(A,B)$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2018 (15 April Evening Shift) PYQ

Solution

JEE MAIN PYQ
$ \text{The integral } \displaystyle \int \left[ \left(\frac{x}{2}\right)^{x} + \left(\frac{2}{x}\right)^{x} \right] \ln!\left(\frac{e x}{2}\right), dx \text{ is equal to:} $





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (8 April Evening Shift) PYQ

Solution

JEE MAIN PYQ
The integral $\displaystyle \int \frac{x^{5}-x^{2}}{(x^{2}+3x+1)\,\tan^{-1}\!\left(x^{3}+\dfrac{1}{x^{2}}\right)}\,dx$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (27 January Evening Shift) PYQ

Solution

JEE MAIN PYQ
The integral $\displaystyle \int \frac{\sin^{2}x \cos^{2}x}{\left(\sin^{5}x + \cos^{3}x \sin^{2}x + \sin^{3}x \cos^{2}x + \cos^{5}x\right)^{2}},dx$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2018 (Offline) PYQ

Solution

JEE MAIN PYQ
The integral $\displaystyle \int \cos(\log_e x)\,dx$ is equal to (where $C$ is a constant of integration):





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (12 January Morning Shift) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle \int \frac{\tan x}{1+\tan x+\tan^{2}x},dx = x - \frac{K}{\sqrt{A}}\tan^{-1}\left(\frac{K\tan x + 1}{\sqrt{A}}\right) + C,\ (C\ \text{is a constant of integration})$ then the ordered pair $(K,A)$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2018 (16 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle \int \frac{\tan x}{1+\tan x+\tan^2 x},dx = x - \frac{K}{\sqrt{A}}\tan^{-1}!\left(\frac{K\tan x + 1}{\sqrt{A}}\right) + C,\ (C\text{ is a constant of integration})$ then the ordered pair $(K,A)$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2018 (16 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle \int \frac{\sin^{2}x+\cos^{2}x}{\sqrt{\sin^{2}x\,\cos^{2}x}\;\sin(x-\theta)}\,dx = A\sqrt{\cos\theta\,\tan x-\sin\theta}\;+\;B\sqrt{\cos\theta-\sin\theta}\,\cot x + C,$ where $C$ is the integration constant, then $AB$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (29 January Evening Shift) PYQ

Solution

JEE MAIN PYQ
For $\alpha, \beta, \gamma, \delta \in \mathbb{N}$, if $\displaystyle \int \left( \left(\dfrac{x}{e}\right)^{2x} + \left(\dfrac{e}{x}\right)^{2x} \right) \log_e x \, dx = \dfrac{1}{\alpha} \left(\dfrac{x}{e}\right)^{\beta x} - \dfrac{1}{\gamma} \left(\dfrac{e}{x}\right)^{\delta x} + C$, where $e = \displaystyle \sum_{n=0}^{\infty} \dfrac{1}{n!}$ and $C$ is the constant of integration, then $\alpha + 2\beta + 3\gamma - 4\delta$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (10 April Evening Shift) PYQ

Solution

JEE MAIN PYQ
Let $I(x) = \int \dfrac{3dx}{(4x+6)\sqrt{4x^2 + 8x + 3}}$ and $I(0) = \dfrac{\sqrt{3}}{4} + 20$. If $I\left(\dfrac{1}{2}\right) = \dfrac{a\sqrt{2}}{b} + c$, where $a,b,c \in \mathbb{N}$, $\gcd(a,b)=1$, then $a + b + c$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

Let $4x+6 = \dfrac{1}{t}$

$I(x) = \frac{3}{4}\sqrt{\frac{4x+2}{4x+6}} + c$

$I(0) = \frac{\sqrt{3}}{4} + c \Rightarrow c = 20$

$I(x) = \frac{3}{4}\sqrt{\frac{4x+2}{4x+6}} + 20$

$I\left(\frac{1}{2}\right) = \frac{3}{4}\sqrt{\frac{8}{8}} + 20 = \frac{3\sqrt{2}}{8} + 20$

$\Rightarrow a + b + c = 3 + 8 + 20 = 31$
JEE MAIN PYQ
The integral $\displaystyle \int \frac{3x^{13}+2x^{11}}{(2x^{4}+3x^{2}+1)^{4}},dx$ is equal to: (where $C$ is a constant of integration)





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (12 January Evening Shift) PYQ

Solution

JEE MAIN PYQ
Let a circle of radius $4$ pass through the origin $O$, the point $A(-\sqrt{3}a,0)$ and $B(0,-\sqrt{2}b)$, where $a$ and $b$ are real parameters and $ab\ne0$. Then the locus of centroid of $\triangle OAB$ is a circle of radius:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

$AB=8$

$3a^2+2b^2=64$

Centroid $G(h,k)$

$h=\frac{-\sqrt{3}a}{3},\quad k=\frac{-\sqrt{2}b}{3}$

$a=-\sqrt{3}h,\quad b=-\frac{3k}{\sqrt{2}}$

Substitute

$9h^2+9k^2=64$

$x^2+y^2=\frac{64}{9}$

$\Rightarrow r=\frac{8}{3}$
JEE MAIN PYQ
Let $I_n=\int \tan^{n}x,dx,\ (n>1)$. If $I_4+I_6=a\tan^{5}x+bx^{5}+C$, where $C$ is a constant of integration, then the ordered pair $(a,b)$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2017 (Offline) PYQ

Solution

JEE MAIN PYQ
$\displaystyle \int \frac{\sin\frac{5x}{2}}{\sin\frac{x}{2}},dx$ is equal to (where $c$ is a constant of integration):





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (8 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
Let $f(x)=\displaystyle \int \frac{2x}{(x^{2}+1)(x^{2}+3)}\,dx$. If $f(3)=\dfrac{1}{2}(\log_{e}5-\log_{e}6)$, then $f(4)$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (25 January Morning Shift) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle \int \frac{\sin^{3/2}x+\cos^{3/2}x}{\sqrt{\sin^2 x,\cos^2 x},\sin(x-\theta)},dx = A\sqrt{\cos\theta,\tan x-\sin\theta}+B\sqrt{\cos\theta-\sin\theta,\cot x}+C,$ where $C$ is the integration constant, then $AB$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (29 January Evening Shift) PYQ

Solution

JEE MAIN PYQ
The integral

$\int {\sqrt {1 + 2\cot x(\cos ecx + \cot x)\,} \,\,dx} $

$\left( {0 < x < {\pi \over 2}} \right)$ is equal to :

(where C is a constant of integration)





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2017 (8 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
If $\int {{{\cos x - \sin x} \over {\sqrt {8 - \sin 2x} }}} dx = a{\sin ^{ - 1}}\left( {{{\sin x + \cos x} \over b}} \right) + c$, where c is a constant of integration, thenthe ordered pair (a, b) is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (24 February Morning Shift) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle \int \frac{dx}{x^{3}(1+x^{6})^{2/3}}=x,f(x),(1+x^{6})^{1/3}+C$ where $C$ is a constant of integration, then the function $f(x)$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (8 April Evening Shift) PYQ

Solution

JEE MAIN PYQ
Let $ \displaystyle \int \frac{2 - \tan x}{3 + \tan x} , dx = \frac{1}{2} \left( \alpha x + \log_e \left| \beta \sin x + \gamma \cos x \right| \right) + C $, where $C$ is the constant of integration. Then $\alpha + \dfrac{\gamma}{\beta}$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (9 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
The integral $\displaystyle \int \sec^{2/3}x , \csc^{4/3}x , dx$ is equal to (Hence $C$ is a constant of integration)





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (9 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle f\left(\frac{3x-4}{3x+4}\right) = x + 2,; x \ne -\frac{4}{3}$ and $\displaystyle \int f(x),dx = A\ln|1-x| + Bx + C,$ then the ordered pair $(A,B)$ is equal to (where $C$ is a constant of integration):





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2017 (9 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
The integral $\int {{1 \over {\root 4 \of {{{(x - 1)}^3}{{(x + 2)}^5}} }}} \,dx$ is equal to : (where C is a constant of integration)





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (31 August Morning Shift) PYQ

Solution

JEE MAIN PYQ
If $ \int \frac{1 - 5\cos^2 x}{\sin^5 x \cos^2 x} dx = f(x) + C $ where $ C $ is the constant of integration, then $ f\left( \frac{\pi}{6} \right) - f\left( \frac{\pi}{4} \right) $ is equal to





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ \int \frac{dx}{\sin^5 x \cos^2 x} - 5 \int \frac{dx}{\sin^5 x} $

$ \int \frac{\sec^2 x dx}{\sin^5 x} - 5 \int \frac{dx}{\sin^5 x} $

By IBP

$ = \frac{\tan x}{\sin^5 x} - \int \frac{\cos x \tan x dx}{\sin^6 x} - 5 \int \frac{dx}{\sin^5 x} $

$ = \frac{\tan x}{\sin^5 x} + C $

$ f(x) = \frac{\tan x}{\sin^5 x} $

$ f\left( \frac{\pi}{6} \right) - f\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{3}} - (\sqrt{2})^5 = 4\sqrt{2} - \frac{32}{\sqrt{3}} $

$ = \frac{4}{\sqrt{3}}(8 - \sqrt{6}) $
JEE MAIN PYQ
Let $ f $ be a polynomial function such that $ f(x^2 + 1) = x^4 + 5x^2 + 2 $, for all $ x \in \mathbb{R} $. Then $ \int_0^3 f(x),dx $ is equal to





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ f(x^2 + 1) = x^4 + 5x^2 + 2 $

put $ x^2 + 1 = t $

$ \Rightarrow f(t) = (t - 1)^2 + 5(t - 1) + 2 $

$ \Rightarrow f(t) = t^2 + 3t - 2 $

Now,

$ \int_0^3 f(t),dt = \int_0^3 (t^2 + 3t - 2),dt $

$ = \left[ \frac{t^3}{3} + \frac{3t^2}{2} - 2t \right]_0^3 $

$ = \frac{27}{3} + \frac{27}{2} - 6 = \frac{33}{2} $
JEE MAIN PYQ
The integral $ \displaystyle \int \frac{2x^{12} + 5x^{9}}{(x^{5} + x^{2} + 1)^{3}}, dx $ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2016 (Offline) PYQ

Solution

JEE MAIN PYQ
If $\int {{{\sin }^{ - 1}}\left( {\sqrt {{x \over {1 + x}}} } \right)} dx$ = A(x)${\tan ^{ - 1}}\left( {\sqrt x } \right)$ + B(x) + C,
where C is a constant of integration, then theordered pair (A(x), B(x)) can be :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 3 September 2020 (Evening) PYQ

Solution

JEE MAIN PYQ
The integral $\int {{{\left( {{x \over {x\sin x + \cos x}}} \right)}^2}dx} $ is equal to
(where C is a constant of integration):





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 4 September 2020 (Morning) PYQ

Solution

JEE MAIN PYQ
Let $f(x) = \int x^3 \sqrt{3 - x^2} , dx.$ If $5f(\sqrt{2}) = -4$, then $f(1)$ is equal to





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (3 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
The integral $\int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{\tan }^3}x.{{\sin }^2}3x\left( {2{{\sec }^2}x.{{\sin }^2}3x + 3\tan x.\sin 6x} \right)dx} $is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 4 September 2020 (Evening) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle \int \frac{dx}{\cos^{3}x\sqrt{2\sin 2x}} = (\tan x)^{A} + C(\tan x)^{B} + k,$ where $k$ is a constant of integration, then $A + B + C$ equals :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2016 (9 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle \int \frac{dx}{(x^{2}-2x+10)^{2}} = A\left(\tan^{-1}\left(\frac{x-1}{3}\right) + \frac{f(x)}{x^{2}-2x+10}\right) + C$ where $C$ is a constant of integration, then:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (10 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
If $f(x)=\displaystyle\int \frac{5x^{8}+7x^{6}}{(x^{2}+1+2x^{7})^{2}}\,dx,\ (x\ge 0)$ and $f(0)=0$, then the value of $f(1)$ is:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (9 January Evening Shift) PYQ

Solution

JEE MAIN PYQ
If$\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} $ = $g\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c$where c is a constant of integration,then g(0) is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 5 September 2020 (Morning) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle \int x^{5}e^{-x^{2}},dx=g(x)e^{-x^{2}}+c$, where $c$ is a constant of integration, then $g(-1)$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (10 April Evening Shift) PYQ

Solution

JEE MAIN PYQ
The integral $\displaystyle \int \frac{dx}{(1+\sqrt{x})\sqrt{x - x^{2}}}$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2016 (10 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle \int e^{x}\!\left(\frac{x\sin^{-1}x}{\sqrt{1-x^{2}}}+\frac{\sin^{-1}x}{(1-x^{2})^{3/2}}+\frac{x}{1-x^{2}}\right)\!dx=g(x)+C$, where $C$ is the constant of integration, then $g\!\left(\dfrac{1}{2}\right)$ equals:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (22 January Evening Shift) PYQ

Solution

JEE MAIN PYQ
Let $n\ge 2$ be a natural number and $0<\theta<\dfrac{\pi}{2}$. Then \[ \int \frac{\big(\sin^{n}\theta-\sin\theta\big)^{1/n}\,\cos\theta}{\sin^{\,n+1}\theta}\,d\theta \] is equal to (where $C$ is a constant of integration):





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (10 January Morning Shift) PYQ

Solution

JEE MAIN PYQ
The integral $\displaystyle \int \dfrac{2x^3 - 1}{x^4 + x} , dx$ is equal to : (Here $C$ is a constant of integration)





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (12 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
The integral $\displaystyle \int \frac{dx}{x^{2}(x^{4}+1)^{3/4}}$ equals :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2015 (Offline) PYQ

Solution

JEE MAIN PYQ
If $\int {{{\cos \theta } \over {5 + 7\sin \theta - 2{{\cos }^2}\theta }}} d\theta $ = A${\log _e}\left| {B\left( \theta \right)} \right| + C$,where C is a constant of integration, then ${{{B\left( \theta \right)} \over A}}$can be :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 5 September 2020 (Evening) PYQ

Solution


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