JEE MAIN Definite Integration Previous Year Questions (PYQs) – Page 1 of 18
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If I1 = $\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}} dx$ andI2 = $\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}} dx$ such that I2 = $\alpha $I1then $\alpha $ equals to :
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Solution
Let $a \in \left(0, \dfrac{\pi}{2}\right)$ be fixed.
If
$\displaystyle \int \dfrac{\tan x + \tan a}{\tan x - \tan a} , dx = A(x)\cos 2a + B(x)\sin 2a + C,$
where $C$ is a constant of integration,
then the functions $A(x)$ and $B(x)$ are respectively:
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Solution
If $I=\displaystyle\int_{0}^{\pi/2}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx$, then $\displaystyle\int_{0}^{2I}\frac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx$ equals:
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Solution
$I=\int_{0}^{\pi/2}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx$
$I+\int_{0}^{\pi/2}\frac{\cos^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx=\frac{\pi}{2}$
⇒ $2I=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4}$
Required integral:
$J=\int_{0}^{2I}\frac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx=\int_{0}^{\pi/2}\frac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx$
Using symmetry:
$2J=\frac{\pi}{2}\int_{0}^{\pi/2}\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx$
Now,
$\int_{0}^{\pi/2}\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx=\frac{\pi}{4}$
So,
$2J=\frac{\pi}{2}\cdot\frac{\pi}{4}=\frac{\pi^2}{8}$
⇒ $J=\frac{\pi^2}{16}$
$\boxed{\frac{\pi^2}{16}}$
$I+\int_{0}^{\pi/2}\frac{\cos^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx=\frac{\pi}{2}$
⇒ $2I=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4}$
Required integral:
$J=\int_{0}^{2I}\frac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx=\int_{0}^{\pi/2}\frac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx$
Using symmetry:
$2J=\frac{\pi}{2}\int_{0}^{\pi/2}\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx$
Now,
$\int_{0}^{\pi/2}\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx=\frac{\pi}{4}$
So,
$2J=\frac{\pi}{2}\cdot\frac{\pi}{4}=\frac{\pi^2}{8}$
⇒ $J=\frac{\pi^2}{16}$
$\boxed{\frac{\pi^2}{16}}$
The value of the definite integral$ \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{dx} \over {(1 + {e^{x\cos x}})({{\sin }^4}x + {{\cos }^4}x)}}} $ is equal to :
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Solution
Let $f(x)=\int_{0}^{x}\left(t+\sin(1-e^{t})\right)dt,\;x\in\mathbb{R}$.
Then, $\displaystyle\lim_{x\to0}\dfrac{f(x)}{x^{3}}$ is equal to:
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Solution
The value of $\displaystyle \int_{-\pi/2}^{\pi/2} \dfrac{dx}{[x] + [\sin x] + 4}$,
where $[t]$ denotes the greatest integer less than or equal to $t$, is :
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Solution
If the area of the bounded region $R = \left\{ {(x,y):\max \{ 0,{{\log }_e}x\} \le y \le {2^x},{1 \over 2} \le x \le 2} \right\}$ is , $\alpha {({\log _e}2)^{ - 1}} + \beta ({\log _e}2) + \gamma $, then the value of ${(\alpha + \beta - 2\lambda )^2}$ is equal to :
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Solution
$\int_{0}^{20\pi} (|\sin x| + |\cos x|)^{2} \, dx$ is equal to:
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Solution
The value of the integral $\displaystyle \int_{-\pi/4}^{\pi/4}\frac{x+\pi/4}{\,2-\cos 2x\,}\,dx$ is:
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Solution
If $\displaystyle \int_{0}^{1} 4\cot^{-1}(1 - 2x + 4x^2),dx = a\tan^{-1}(2) - b\log_e(5)$, where $a,b \in \mathbb{N}$, then $(2a + b)$ is equal to ______.
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Solution
Let $I = \int_{0}^{1} \cot^{-1}(1 - 2x + 4x^2),dx$
$I = \int_{0}^{1} \left(\cot^{-1}(2x-1) - \cot^{-1}(2x)\right),dx \quad ...(1)$
Applying King
$I = \int_{0}^{1} \left(-\cot^{-1}(2x-1) + \cot^{-1}(2x-2)\right),dx \quad ...(2)$
From (1) & (2)
$2I = \int_{0}^{1} \left(\cot^{-1}(2x-2) - \cot^{-1}(2x)\right),dx$
$= \int_{0}^{1} \cot^{-1}(2x-2),dx - \int_{0}^{1} \cot^{-1}(2x),dx$
Applying King
$= \int_{0}^{1} \cot^{-1}(-2x),dx - \int_{0}^{1} \cot^{-1}(2x),dx$
$= \int_{0}^{1} (\pi - \cot^{-1}(2x)),dx - \int_{0}^{1} \cot^{-1}(2x),dx$
$= \int_{0}^{1} \pi,dx - 2\int_{0}^{1} \cot^{-1}(2x),dx$
$= \pi - 2\int_{0}^{1} \cot^{-1}(2x),dx$
By parts
$I = \pi - 2\left[x\cot^{-1}(2x)\right]{0}^{1} + \int{0}^{1} \frac{2x}{1+4x^2},dx$
Let $1 + 4x^2 = t$
$8x,dx = dt$
$I = \pi - 2\cot^{-1}(2) + \frac{1}{4}\int_{1}^{5} \frac{dt}{t}$
$= \pi - 2\cot^{-1}(2) + \frac{1}{4}\ln 5$
$\Rightarrow 2I = 2\pi - 4\cot^{-1}(2) + \frac{1}{2}\ln 5$
Given $\int_{0}^{1} 4\cot^{-1}(1 - 2x + 4x^2),dx = 4I$
$= 2\left[2\pi - 4\cot^{-1}(2) + \frac{1}{2}\ln 5\right]$
$= 4\pi - 8\cot^{-1}(2) + \ln 5$
$\Rightarrow 2a + b = 8 + 1 = 9$
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