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Phrases Previous Year Questions (PYQs)
Phrases Definite Integration PYQ
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2009 PYQ
The value of
$\displaystyle \int_{0}^{\pi} \frac{x \sin x}{1+\cos^2 x},dx$
is:
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Phrases Previous Year PYQ Phrases NIMCET 2009 PYQ
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,
$I = \dfrac{\pi^2}{4}$
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2012 PYQ
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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Phrases Previous Year PYQ Phrases NIMCET 2012 PYQ
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 $
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2012 PYQ
The value of integral $\displaystyle \int_{0}^{\pi/2} \log \tan x dx$ is
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Phrases Previous Year PYQ Phrases NIMCET 2012 PYQ
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: $0$
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2012 PYQ
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:
Go to Discussion
Phrases Previous Year PYQ Phrases NIMCET 2012 PYQ
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:
$I = \dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4}$
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2011 PYQ
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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Phrases Previous Year PYQ Phrases NIMCET 2011 PYQ
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 $
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2011 PYQ
$ \displaystyle \int_{0}^{1/2} \frac{dx}{\sqrt{x - x^{2}}} $
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Phrases Previous Year PYQ Phrases NIMCET 2011 PYQ
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} $
then x =
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2013 PYQ
If $I_n = \int_0^{\pi/4} tan^{n} \theta d\theta$ , then $I_8 + I_6$ equals
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Phrases Previous Year PYQ Phrases NIMCET 2013 PYQ
Solution
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2013 PYQ
The value of the integral $\int _0^{\pi/2} \frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx$ is
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Phrases Previous Year PYQ Phrases NIMCET 2013 PYQ
Solution
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2019 PYQ
Not Available
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Phrases Previous Year PYQ Phrases NIMCET 2019 PYQ
Solution
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2022 PYQ
If $a{\lt}b$ then $\int ^b_a\Bigg{(}|x-a|+|x-b|\Bigg{)}dx$ is equal to
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Phrases Previous Year PYQ Phrases NIMCET 2022 PYQ
Solution
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2024 PYQ
Which of the following is TRUE?
A. If $f$ is continuous on $[a,b]$, then $\int ^b_axf(x)\mathrm{d}x=x\int ^b_af(x)\mathrm{d}x$
B. $\int ^3_0{e}^{{x}^2}dx=\int ^5_0e^{{x}^2}dx+{\int ^5_3e}^{{x}^2}dx$
C. If $f$ is continuous on $[a,b]$, then $\frac{d}{\mathrm{d}x}\Bigg{(}\int ^b_af(x)dx\Bigg{)}=f(x)$
D. Both (a) and (b)
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Phrases Previous Year PYQ Phrases NIMCET 2024 PYQ
Solution
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2024 PYQ
If for non-zero x, $cf(x)+df\Bigg{(}\frac{1}{x}\Bigg{)}=|\log |x||+3,$ where $c\ne 0$, then $\int ^e_1f(x)dx=$
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Phrases Previous Year PYQ Phrases NIMCET 2024 PYQ
Solution
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2008 PYQ
The value of
$\displaystyle \int_0^{\pi/2} \frac{dx}{1+\tan^3 x}$
is
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Phrases Previous Year PYQ Phrases NIMCET 2008 PYQ
Solution
Using property $\int_0^{\pi/2} f(\tan x)dx = \int_0^{\pi/2} f(\cot x),dx$Add both:
$I=\int_0^{\pi/2} \frac{1}{1+\tan^3 x}dx$
$I=\int_0^{\pi/2} \frac{\tan^3 x}{1+\tan^3 x}dx$
$2I=\int_0^{\pi/2}1dx=\frac{\pi}{2}$
$I=\frac{\pi}{4}$
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2008 PYQ
If $f:\mathbb R\to\mathbb R$ and $g:\mathbb R\to\mathbb R$ are continuous functions, then evaluate
$\displaystyle \int_{-\pi/2}^{\pi/2}[f(x)+f(-x)][g(x)-g(-x)],dx$
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Phrases Previous Year PYQ Phrases NIMCET 2008 PYQ
Solution
$f(x)+f(-x)$ is an even function $g(x)-g(-x)$ is an odd function Product of even and odd function is odd Integral of odd function over symmetric limits is $0$
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2018 PYQ
$\int_0^{\pi} x\, f(\sin x)\, dx$ is equal to
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Phrases Previous Year PYQ Phrases NIMCET 2018 PYQ
Solution
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2018 PYQ
Let $f: R \rightarrow R$ be defined by $f(x) = \begin{cases} x + 2 & \text{if } x < 0 \\ |x - 2| & \text{if } x \geq 0 \end{cases}$. Find $\int_{-2}^{3} f(x)\, dx$
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Phrases Previous Year PYQ Phrases NIMCET 2018 PYQ
Solution
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2025 PYQ
The value of $\frac{d}{dx}\int ^{2\sin x}_{\sin {x}^2}{e}^{{t}^2}dt$ at $x=\pi$
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Phrases Previous Year PYQ Phrases NIMCET 2025 PYQ
Solution
We want to find $\dfrac{d}{dx}\left(\int_{\sin^2 x}^{2\sin x} e^{t^2}, dt\right)$ at $x=\pi$.Using Leibniz rule:
$\dfrac{d}{dx}\left(\int_{a(x)}^{b(x)} f(t) dt\right)
= f(b(x)) b'(x) ;-; f(a(x)) a'(x)$
Here
$a(x)=\sin^2 x$,
$b(x)=2\sin x$,
$f(t)=e^{t^2}$.
Compute derivatives:
$a'(x)=2\sin x\cos x$
$b'(x)=2\cos x$
So the derivative is:
$e^{(2\sin x)^2}(2\cos x)-e^{(\sin^2 x)^2}(2\sin x\cos x)$
Now evaluate at $x=\pi$
$\sin\pi=0,\quad \cos\pi=-1$
Thus:
$b'(\pi)=2(-1)=-2$
$a'(\pi)=0$
Therefore:
$e^{0}(-2)-e^{0}(0)=-2$
So the final answer is:
$-2$
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2010 PYQ
The value of the integral
$\displaystyle \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}}, dx$ is
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Phrases Previous Year PYQ Phrases NIMCET 2010 PYQ
Solution
Use substitution: x → 9 – x I = ∫ √x / (√(9–x) + √x) dx I = ∫ √(9–x) / (√x + √(9–x)) dx Add both expressions: 2I = ∫₃⁶ 1 dx = 3 ⇒ I = 3/2
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2010 PYQ
The value of the integral
$\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{3 + \sin 2x}, dx$ is
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Phrases Previous Year PYQ Phrases NIMCET 2010 PYQ
Solution
sin2x = 2 sinx cosx Rewrite numerator: sinx + cosx = √2 sin(x + π/4) Denominator: 3 + sin2x = 3 + 2 sinx cosx = 1 + (sinx + cosx)² Let t = sinx + cosx dt/dx = cosx – sinx → use identity to convert dx Integral evaluates to (1/4) log 3
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2010 PYQ
$I_1 = \int_{0}^{1} 2x^2 dx,\
I_2 = \int_{0}^{1} 2x^3 dx,\
I_3 = \int_{1}^{2} x^2 dx,\
I_4 = \int_{1}^{2} 2x^3 dx$
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Phrases Previous Year PYQ Phrases NIMCET 2010 PYQ
Solution
$I_1 = \left[\frac{2x^3}{3}\right]_{0}^{1} = \frac{2}{3}$ $I_2 = \left[\frac{2x^4}{4}\right]_{0}^{1} = \frac{1}{2}$ Thus $I_1 > I_2$ $I_3 = \left[\frac{x^3}{3}\right]_{1}^{2} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$ $I_4 = \left[\frac{2x^4}{4}\right]_{1}^{2} = 4 - \frac{1}{2} = \frac{7}{2}$ Thus $I_4 > I_3$
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2018 PYQ
If 
and 
, then the value of 
is
and , then the value of isGo to Discussion
Phrases Previous Year PYQ Phrases NIMCET 2018 PYQ
Solution
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2014 PYQ
The value of $\int_{0}^{\pi/4} log(1+tanx)dx$ is equal to:
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Phrases Previous Year PYQ Phrases NIMCET 2014 PYQ
Solution
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2014 PYQ
If [x] represents the greatest integer not exceeding x, then $\int_{0}^{9} [x] dx $ is
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Phrases Previous Year PYQ Phrases NIMCET 2014 PYQ
Solution
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2018 PYQ
Through any point (x, y) of a curve which passes through the origin, lines are drawn parallel to the coordinate axes. The curve, given that it divides the rectangle formed by the two lines and the axes into two areas, one of which is twice the other, represents a family of
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Phrases Previous Year PYQ Phrases NIMCET 2018 PYQ
Solution
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2017 PYQ
The value of $\int_{0}^{\pi}x^3 \sin x dx$
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Phrases Previous Year PYQ Phrases NIMCET 2017 PYQ
Solution
Let $I=\displaystyle\int_{0}^{\pi}x^{3}\sin x\,dx$.
Using integration by parts: let $u=x^{3},\; dv=\sin x\,dx$ $\Rightarrow du=3x^{2}dx,\; v=-\cos x$
$I = [-x^{3}\cos x]_0^{\pi} + \displaystyle\int_{0}^{\pi}3x^{2}\cos x\,dx$
Now, let $J=\displaystyle\int_{0}^{\pi}x^{2}\cos x\,dx$
Again, by parts: $u=x^{2},\; dv=\cos x\,dx \Rightarrow du=2x\,dx,\; v=\sin x$
$J = [x^{2}\sin x]_0^{\pi} - \displaystyle\int_{0}^{\pi}2x\sin x\,dx$
Let $K=\displaystyle\int_{0}^{\pi}x\sin x\,dx$ By parts: $u=x,\; dv=\sin x\,dx \Rightarrow du=dx,\; v=-\cos x$
$K=[-x\cos x]_0^{\pi}+\displaystyle\int_{0}^{\pi}\cos x\,dx = \pi$
So $J=0-2K=-2\pi$
Then $\displaystyle\int_{0}^{\pi}3x^{2}\cos x\,dx = 3J = -6\pi$
$I = [-x^{3}\cos x]_0^{\pi} + 3J = (-\pi^{3}\cos\pi - 0) - 6\pi = \pi^{3} - 6\pi$
Answer: $\boxed{\pi^{3} - 6\pi}$ ✅
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2017 PYQ
Evaluate $\displaystyle \int_{0}^{1}x(1-x)^ndx $
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Phrases Previous Year PYQ Phrases NIMCET 2017 PYQ
Solution
Let $t=1-x \Rightarrow x=1-t,\ dx=-dt$.
As $x:0\to1$,
$t:1\to0$.
\[ \int_{0}^{1} x(1-x)^n\,dx \]
\[ = \int_{1}^{0} (1-t)\,t^n\,(-dt) \]
\[= \int_{0}^{1} \big(t^n - t^{n+1}\big)\,dt \]
\[ = \left[\frac{t^{n+1}}{n+1}-\frac{t^{n+2}}{n+2}\right]_{0}^{1} = \frac{1}{n+1}-\frac{1}{n+2}. \]
Simplify: \[ \frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)}. \]
Answer: $\boxed{\dfrac{1}{(n+1)(n+2)}}$ (valid for $n>-1$).
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2015 PYQ
The value of $\int_{-\pi/3}^{\pi/3} \frac{x sinx}{cos^{2}x}dx$
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Phrases Previous Year PYQ Phrases NIMCET 2015 PYQ
Solution
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2025 PYQ
The value of $\int ^{\frac{\pi}{2}}_0\frac{(1+2\cos x)}{({2+\cos x)}^2}dx$ lies in the interval
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Phrases Previous Year PYQ Phrases NIMCET 2025 PYQ
Solution
We need to evaluate $ \displaystyle \int_{0}^{\pi/2} \frac{1 + 2\cos x}{(2 + \cos x)^2}\, dx $.Let $ t = 2 + \cos x $.
Then
$ dt = -\sin x\, dx $.
But the integral has no $\sin x$, so rewrite numerator:
$ 1 + 2\cos x = (2 + \cos x) - 1 = t - 1 $.
Now express $ dx $ using
$ \sin^2 x = 1 - \cos^2 x $,
but the standard trick is to differentiate:
$ \dfrac{d}{dx}\left(\dfrac{1}{2+\cos x}\right)
= -\dfrac{-\sin x}{(2+\cos x)^2}
= \dfrac{\sin x}{(2+\cos x)^2}. $
We
use complementary substitution:
Let $ x = \frac{\pi}{2} - y $.
Then $\cos x = \sin y$ and $\sin x = \cos y$.
Integral becomes:
$ I = \int_{0}^{\pi/2} \frac{1 + 2\sin y}{(2 + \sin y)^2}\, dy. $
Average the two forms:
$ I = \frac{1}{2}\int_{0}^{\pi/2}
\left[
\frac{1 + 2\cos x}{(2 + \cos x)^2} +
\frac{1 + 2\sin x}{(2 + \sin x)^2}
\right] dx. $
Now observe identity:
$ \frac{1 + 2\cos x}{(2 + \cos x)^2}
+ \frac{1 + 2\sin x}{(2 + \sin x)^2}
= \frac{d}{dx}\left(\frac{\sin x - \cos x}{(2+\cos x)(2+\sin x)}\right). $
Thus integral becomes a telescoping form and evaluates to:
$ I = \left[ \frac{\sin x - \cos x}{(2+\cos x)(2+\sin x)} \right]_{0}^{\pi/2}. $
Now compute:
At $ x = \frac{\pi}{2}$:
$ \sin x = 1,\;\cos x = 0 $
Expression =
$ \dfrac{1 - 0}{(2+0)(2+1)} = \dfrac{1}{6}. $
At $ x = 0$:
$ \sin 0 = 0,\;\cos 0 = 1 $
Expression =
$ \dfrac{0 - 1}{(2+1)(2+0)} = -\dfrac{1}{6}. $
Therefore:
$ I = \dfrac{1}{6} - (-\dfrac{1}{6}) = \dfrac{2}{6} = \dfrac{1}{3}. $
Final Answer: $\dfrac{1}{3} $
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2020 PYQ
If 
where n is a positive
integer, then the relation between In and In-1 is
where n is a positive
integer, then the relation between In and In-1 isGo to Discussion
Phrases Previous Year PYQ Phrases NIMCET 2020 PYQ
Solution
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2020 PYQ
The value of 
depends on the
depends on theGo to Discussion
Phrases Previous Year PYQ Phrases NIMCET 2020 PYQ
Solution
Phrases PYQ
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Phrases Previous Year PYQ Phrases NIMCET 2015 PYQ
$\int_0^\pi [cotx]dx$ where [.] denotes the greatest integer function, is equal to
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Phrases Previous Year PYQ Phrases NIMCET 2015 PYQ
Solution
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Phrases Previous Year PYQ Phrases NIMCET 2023 PYQ
The value of $\int ^{\pi/3}_{-\pi/3}\frac{x\sin x}{{\cos }^2x}dx$ is
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Phrases Previous Year PYQ Phrases NIMCET 2023 PYQ
Solution
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