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NIMCET 2027 Previous Year Questions (PYQs)

NIMCET 2027 Definite Integration PYQ

NIMCET 2027 PYQ
The value of $\displaystyle \int_{0}^{\pi} \frac{x \sin x}{1+\cos^2 x},dx$ is:





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NIMCET 2027 Previous Year PYQ NIMCET 2027 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, $I = \dfrac{\pi^2}{4}$
NIMCET 2027 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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NIMCET 2027 Previous Year PYQ NIMCET 2027 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 $
NIMCET 2027 PYQ
The value of integral $\displaystyle \int_{0}^{\pi/2} \log \tan x dx$ is





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NIMCET 2027 Previous Year PYQ NIMCET 2027 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: $0$
NIMCET 2027 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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NIMCET 2027 Previous Year PYQ NIMCET 2027 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: $I = \dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4}$
NIMCET 2027 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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NIMCET 2027 Previous Year PYQ NIMCET 2027 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 $
NIMCET 2027 PYQ
$ \displaystyle \int_{0}^{1/2} \frac{dx}{\sqrt{x - x^{2}}} $





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NIMCET 2027 Previous Year PYQ NIMCET 2027 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} $
NIMCET 2027 PYQ
If \(\displaystyle \int \limits_{\log 2}^{x} \frac{1}{\sqrt{e^{x} - 1}} \, dt = \dfrac{\pi}{6}\), then \(x =\)





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NIMCET 2027 Previous Year PYQ NIMCET 2027 PYQ

Solution

NIMCET 2027 PYQ
If $I_n = \int_0^{\pi/4} tan^{n} \theta d\theta$ , then $I_8 + I_6$ equals





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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2013 PYQ

Solution

NIMCET 2027 PYQ
The value of the integral $\int _0^{\pi/2} \frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx$ is





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Solution

NIMCET 2027 PYQ
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 PYQ
If $a{\lt}b$ then $\int ^b_a\Bigg{(}|x-a|+|x-b|\Bigg{)}dx$ is equal to





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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2022 PYQ

Solution

NIMCET 2027 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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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2024 PYQ

Solution

NIMCET 2027 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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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2024 PYQ

Solution

NIMCET 2027 PYQ
The value of $\displaystyle \int_0^{\pi/2} \frac{dx}{1+\tan^3 x}$ is





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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2008 PYQ

Solution

Using property $\int_0^{\pi/2} f(\tan x)dx = \int_0^{\pi/2} f(\cot x),dx$ 
Add both: $I=\int_0^{\pi/2} \frac{1}{1+\tan^3 x}dx$ 
$I=\int_0^{\pi/2} \frac{\tan^3 x}{1+\tan^3 x}dx$ 
$2I=\int_0^{\pi/2}1dx=\frac{\pi}{2}$ 
$I=\frac{\pi}{4}$
NIMCET 2027 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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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 $0$
NIMCET 2027 PYQ
$\int_0^{\pi} x\, f(\sin x)\, dx$ is equal to





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Solution

NIMCET 2027 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

NIMCET 2027 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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NIMCET 2027 Previous Year PYQ NIMCET 2027 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: $-2$
NIMCET 2027 PYQ
The value of the integral $\displaystyle \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}}, dx$ is





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NIMCET 2027 Previous Year PYQ NIMCET 2027 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
NIMCET 2027 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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NIMCET 2027 Previous Year PYQ NIMCET 2027 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
NIMCET 2027 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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NIMCET 2027 Previous Year PYQ NIMCET 2027 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 $I_4 > I_3$
NIMCET 2027 PYQ
If  and , then the value of  is







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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2018 PYQ

Solution

NIMCET 2027 PYQ
The value of $\int_{0}^{\pi/4} log(1+tanx)dx$ is equal to:





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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2014 PYQ

Solution

NIMCET 2027 PYQ
If [x] represents the greatest integer not exceeding x, then $\int_{0}^{9} [x] dx $ is





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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2014 PYQ

Solution

NIMCET 2027 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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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2018 PYQ

Solution

NIMCET 2027 PYQ
The value of $\int_{0}^{\pi}x^3 \sin x dx$





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NIMCET 2027 Previous Year PYQ NIMCET 2027 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: $\boxed{\pi^{3} - 6\pi}$ ✅

NIMCET 2027 PYQ
Evaluate $\displaystyle \int_{0}^{1}x(1-x)^ndx $





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NIMCET 2027 Previous Year PYQ NIMCET 2027 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 $n>-1$).

NIMCET 2027 PYQ
The value of $\int_{-\pi/3}^{\pi/3} \frac{x sinx}{cos^{2}x}dx$





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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2015 PYQ

Solution

NIMCET 2027 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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NIMCET 2027 Previous Year PYQ NIMCET 2027 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} $
NIMCET 2027 PYQ
If  where n is a positive integer, then the relation between In and In-1 is





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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2020 PYQ

Solution

NIMCET 2027 PYQ
The value of  depends on the





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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2020 PYQ

Solution

NIMCET 2027 PYQ
$\int_0^\pi [cotx]dx$ where [.] denotes the greatest integer function, is equal to





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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2015 PYQ

Solution

NIMCET 2027 PYQ
The value of $\int ^{\pi/3}_{-\pi/3}\frac{x\sin x}{{\cos }^2x}dx$ is





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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2023 PYQ

Solution

NIMCET 2027 PYQ
Evaluate the following definite integral:$\int_{0}^{\frac{\pi}{4}} \frac{dx}{\cos^4 x}$





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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2026 PYQ

Solution

$\int_{0}^{\frac{\pi}{4}} \frac{dx}{\cos^4 x}$ 
$=\int_{0}^{\frac{\pi}{4}} \sec^4 xdx$ 
Now, $\sec^4 x=\sec^2 x(1+\tan^2 x)$ 

Let $\tan x=t$ $\sec^2 xdx=dt$ 

Limits: 
$x=0 \Rightarrow t=0$ 
$x=\frac{\pi}{4} \Rightarrow t=1$ 

So, $\int_0^1 (1+t^2)dt$ 
$=\left[t+\frac{t^3}{3}\right]_0^1$ 
$=1+\frac{1}{3}$ 
$=\frac{4}{3}$
NIMCET 2027 PYQ
If $f:[0,\infty)\to R$ is defined by $f(x)=\frac{x^2-1}{x^2+1}$, find the value of: $\int_{-1}^{1} f^{-1}(y)dy$





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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2026 PYQ

Solution

Let 
$y=\frac{x^2-1}{x^2+1}$ 
$y(x^2+1)=x^2-1$ 
$yx^2+y=x^2-1$ 
$x^2(y-1)=-(1+y)$ 
$x^2=\frac{1+y}{1-y}$ 

Since $x\ge 0$, 
$f^{-1}(y)=\sqrt{\frac{1+y}{1-y}}$ 

Now, $I=\int_{-1}^{1}\sqrt{\frac{1+y}{1-y}}dy$ 
Put 
$y=\cos\theta$ 
$dy=-\sin\theta,d\theta$ 

When $y=-1,\ \theta=\pi$ 
When $y=1,\ \theta=0$ 

$I=\int_{\pi}^{0}\sqrt{\frac{1+\cos\theta}{1-\cos\theta}}(-\sin\theta)d\theta$ 

$=\int_0^\pi \cot\frac{\theta}{2}\sin\theta d\theta$ 

Now, $\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$ 
$I=\int_0^\pi 2\cos^2\frac{\theta}{2} d\theta$ 
$=\int_0^\pi (1+\cos\theta) d\theta$ 
$=[\theta+\sin\theta]_0^\pi$ 
$=\pi$ 
Answer: $\pi$

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