NIMCET 2027 Definite Integration Previous Year Questions (PYQs) – Page 1 of 4
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Showing 35 questions
The value of
$\displaystyle \int_{0}^{\pi} \frac{x \sin x}{1+\cos^2 x},dx$
is:
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Solution
Let
$I = \displaystyle \int_{0}^{\pi} \frac{x \sin x}{1+\cos^2 x}dx$
Use property:
$I = \int_{0}^{\pi} f(x)dx = \int_{0}^{\pi} f(\pi - x)dx$
Compute
$f(\pi - x) = \dfrac{(\pi - x)\sin(\pi - x)}{1 + \cos^2(\pi - x)}
= \dfrac{(\pi - x)\sin x}{1 + \cos^2 x}$
Add them:
$f(x) + f(\pi - x) = \dfrac{\pi \sin x}{1 + \cos^2 x}$
So,
$2I = \displaystyle \int_{0}^{\pi} \frac{\pi \sin x}{1 + \cos^2 x}dx$
Let $u = \cos x$, $du = -\sin x,dx$.
When $x=0$, $u=1$, and when $x=\pi$, $u=-1$:
$2I = \pi \displaystyle \int_{1}^{-1} \frac{-du}{1+u^2}$
$2I = \pi \displaystyle \int_{-1}^{1} \frac{du}{1+u^2}$
This equals:
$2I = \pi\left[\tan^{-1}u\right]_{-1}^{1}
= \pi\left(\dfrac{\pi}{4} - \left(-\dfrac{\pi}{4}\right)\right)$
$2I = \pi \cdot \dfrac{\pi}{2} = \dfrac{\pi^2}{2}$
So,
$I = \dfrac{\pi^2}{4}$
If
$ I_1 = \displaystyle \int_{0}^{1} 2^{x^2},dx,\quad
I_2 = \displaystyle \int_{0}^{1} 2^{x^3},dx,\quad
I_3 = \displaystyle \int_{1}^{2} 2^{x^2},dx,\quad
I_4 = \displaystyle \int_{1}^{2} 2^{x^3},dx,$
then
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Solution
On the interval $0 \le x \le 1$:
$ x^3 < x^2 \Rightarrow 2^{x^3} < 2^{x^2} $
So,
$ I_2 = \displaystyle \int_0^1 2^{x^3},dx < \int_0^1 2^{x^2}dx = I_1 $
Thus,
$ I_1 > I_2 $
On the interval $1 \le x \le 2$:
$ x^3 > x^2 \Rightarrow 2^{x^3} > 2^{x^2} $
So,
$ I_4 = \displaystyle \int_1^2 2^{x^3}dx > \int_1^2 2^{x^2}dx = I_3 $
Thus,
$ I_4 > I_3 $
The value of integral $\displaystyle \int_{0}^{\pi/2} \log \tan x dx$ is
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Solution
Let $I = \displaystyle \int_{0}^{\pi/2} \log \tan x dx$
Using property $\displaystyle \int_{0}^{\pi/2} f(x)dx = \int_{0}^{\pi/2} f\left(\dfrac{\pi}{2}-x\right)dx$
$I = \displaystyle \int_{0}^{\pi/2} \log \tan\left(\dfrac{\pi}{2}-x\right) dx$
$= \displaystyle \int_{0}^{\pi/2} \log \cot x dx$
$= \displaystyle \int_{0}^{\pi/2} \log\left(\dfrac{1}{\tan x}\right) dx$
$= \displaystyle \int_{0}^{\pi/2} [-\log \tan x] dx$
$I= -I$
$\Rightarrow I + I = 0 \Rightarrow 2I = 0 \Rightarrow I = 0$
Answer: $0$
The value of
$\displaystyle \int_{0}^{\sin^2 x} \sin^{-1}\sqrt{t} dt + \int_{0}^{\cos^2 x} \cos^{-1}\sqrt{t} dt$ is:
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Solution
Use the identity:
$\sin^{-1}y+\cos^{-1}y=\dfrac{\pi}{2}$
So:
$\cos^{-1}\sqrt{t}=\dfrac{\pi}{2}-\sin^{-1}\sqrt{t}$
Now substitute:
$I=\displaystyle\int_0^{\sin^2 x}\sin^{-1}\sqrt{t} dt + \int_0^{\cos^2 x} \left(\dfrac{\pi}{2}-\sin^{-1}\sqrt{t}\right) dt$
$I=\dfrac{\pi}{2}\cos^2 x + \int_0^{\sin^2 x}\sin^{-1}\sqrt{t}dt - \int_0^{\cos^2 x}\sin^{-1}\sqrt{t} dt$
Combine integrals:
$I=\dfrac{\pi}{2}\cos^2 x + \int_{\cos^2 x}^{\sin^2 x}\sin^{-1}\sqrt{t} dt$
But:
$\sin^2 x + \cos^2 x = 1$
Limits become from $1$ to $0$:
$I=\dfrac{\pi}{2}(1 - \sin^2 x) + \int_{1}^{0}\sin^{-1}\sqrt{t} dt$
$I=\dfrac{\pi}{2} - \int_0^{1}\sin^{-1}\sqrt{t} dt$
Let $u=\sqrt{t}$,
$dt=2udu$:
$\displaystyle \int_0^{1}\sin^{-1}\sqrt{t} dt = 2\int_0^{1} u\sin^{-1}udu$
Standard result:
$\displaystyle 2\int_0^{1} u\sin^{-1}u du = \dfrac{\pi}{4}$
Thus:
$I = \dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4}$
If $a$ is a positive integer, then the number of values satisfying
$ \displaystyle \int_{0}^{\pi/2} \left[ a^{2}\left(\frac{\cos 3x}{4}+\frac{3}{4}\cos x\right)+a\sin x - 20\cos x \right] dx \le -\frac{a^{2}}{3} $
is
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Solution
$ \displaystyle \int_{0}^{\pi/2} \cos 3x, dx = \frac{1}{3},;;
\int_{0}^{\pi/2} \cos x, dx = 1,;;
\int_{0}^{\pi/2} \sin x, dx = 1 $
So integral becomes
$ \displaystyle a^{2}\left(\frac{1}{12}+\frac{3}{4}\right)+a - 20
= \frac{5a^{2}}{6} + a - 20 $
Given
$ \displaystyle \frac{5a^{2}}{6}+a-20 \le -\frac{a^{2}}{3} $
$ \displaystyle \Rightarrow \frac{7a^{2}}{6} + a - 20 \le 0 $
Multiply by 6:
$ 7a^{2} + 6a - 120 \le 0 $
Roots:
$ a = \frac{26}{7} \approx 3.714 $
So valid positive integers:
$ a = 1,,2,,3 $
$ \displaystyle \int_{0}^{1/2} \frac{dx}{\sqrt{x - x^{2}}} $
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Solution
$ x = \sin^{2}\theta \Rightarrow dx = 2\sin\theta\cos\theta, d\theta $
$ \sqrt{x-x^{2}} = \sin\theta\cos\theta $
Integral becomes $ \int 2, d\theta $
Limits: $0 \to 0$, $\frac12 \to \frac{\pi}{4}$
Value $ = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} $
If \(\displaystyle \int \limits_{\log 2}^{x} \frac{1}{\sqrt{e^{x} - 1}} \, dt = \dfrac{\pi}{6}\), then \(x =\)
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Solution
If $I_n = \int_0^{\pi/4} tan^{n} \theta d\theta$ , then $I_8 + I_6$ equals
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Solution
The value of the integral $\int _0^{\pi/2} \frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx$ is
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Solution
Let f(x) be a polynomial satisfying \(f(0) = 2\), \(f'(0) = 3\) and \(f''(x) = f(x)\). Then \(f(4)\) is equal to:
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Solution
NIMCET 2027
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