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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Definite Integration PYQ

MCA NIMCET PYQ
The value of $\displaystyle \int_{0}^{\pi} \frac{x \sin x}{1+\cos^2 x},dx$ is:





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MCA NIMCET Previous Year PYQ MCA NIMCET 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, $I = \dfrac{\pi^2}{4}$
MCA NIMCET 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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MCA NIMCET Previous Year PYQ MCA NIMCET 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 $
MCA NIMCET PYQ
The value of integral $\displaystyle \int_{0}^{\pi/2} \log \tan x dx$ is





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MCA NIMCET Previous Year PYQ MCA NIMCET 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: $0$
MCA NIMCET 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:





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MCA NIMCET Previous Year PYQ MCA NIMCET 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: $I = \dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4}$
MCA NIMCET 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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MCA NIMCET Previous Year PYQ MCA NIMCET 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 $
MCA NIMCET PYQ
$ \displaystyle \int_{0}^{1/2} \frac{dx}{\sqrt{x - x^{2}}} $





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MCA NIMCET Previous Year PYQ MCA NIMCET 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} $
MCA NIMCET PYQ
If  then x =





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

MCA NIMCET PYQ
If $I_n = \int_0^{\pi/4} tan^{n} \theta d\theta$ , then $I_8 + I_6$ equals





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

MCA NIMCET PYQ
The value of the integral $\int _0^{\pi/2} \frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx$ is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

MCA NIMCET PYQ
If $a{\lt}b$ then $\int ^b_a\Bigg{(}|x-a|+|x-b|\Bigg{)}dx$ is equal to





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

MCA NIMCET 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

MCA NIMCET 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

MCA NIMCET PYQ
The value of $\displaystyle \int_0^{\pi/2} \frac{dx}{1+\tan^3 x}$ is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

Using property $\int_0^{\pi/2} f(\tan x)dx = \int_0^{\pi/2} f(\cot x),dx$ 
Add both: $I=\int_0^{\pi/2} \frac{1}{1+\tan^3 x}dx$ 
$I=\int_0^{\pi/2} \frac{\tan^3 x}{1+\tan^3 x}dx$ 
$2I=\int_0^{\pi/2}1dx=\frac{\pi}{2}$ 
$I=\frac{\pi}{4}$
MCA NIMCET 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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MCA NIMCET Previous Year PYQ MCA NIMCET 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 $0$
MCA NIMCET PYQ
$\int_0^{\pi} x\, f(\sin x)\, dx$ is equal to





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

MCA NIMCET 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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Solution

MCA NIMCET 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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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: $-2$
MCA NIMCET PYQ
The value of the integral $\displaystyle \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}}, dx$ is





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MCA NIMCET Previous Year PYQ MCA NIMCET 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
MCA NIMCET 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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MCA NIMCET Previous Year PYQ MCA NIMCET 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
MCA NIMCET 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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MCA NIMCET Previous Year PYQ MCA NIMCET 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 $I_4 > I_3$
MCA NIMCET PYQ
If  and , then the value of  is







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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

MCA NIMCET PYQ
The value of $\int_{0}^{\pi/4} log(1+tanx)dx$ is equal to:





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

MCA NIMCET PYQ
If [x] represents the greatest integer not exceeding x, then $\int_{0}^{9} [x] dx $ is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

MCA NIMCET 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

MCA NIMCET PYQ
The value of $\int_{0}^{\pi}x^3 \sin x dx$





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MCA NIMCET Previous Year PYQ MCA NIMCET 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: $\boxed{\pi^{3} - 6\pi}$ ✅

MCA NIMCET PYQ
Evaluate $\displaystyle \int_{0}^{1}x(1-x)^ndx $





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MCA NIMCET Previous Year PYQ MCA NIMCET 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 $n>-1$).

MCA NIMCET PYQ
The value of $\int_{-\pi/3}^{\pi/3} \frac{x sinx}{cos^{2}x}dx$





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

MCA NIMCET 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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MCA NIMCET Previous Year PYQ MCA NIMCET 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} $
MCA NIMCET PYQ
If  where n is a positive integer, then the relation between In and In-1 is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

MCA NIMCET PYQ
The value of  depends on the





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

MCA NIMCET PYQ
$\int_0^\pi [cotx]dx$ where [.] denotes the greatest integer function, is equal to





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

MCA NIMCET PYQ
The value of $\int ^{\pi/3}_{-\pi/3}\frac{x\sin x}{{\cos }^2x}dx$ is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution


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