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CGPET 2021 Solved Paper

Section: Mathematics

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Question : 111 of 150

Marks: +1, -0

The value of

\int\limits_{0}^{\pi/2} \; \frac{dx}{1+\cot x}

$\; \frac{\pi}{2}$
$\; \frac{\pi}{4}$
$\; \frac{1}{2}$
1

Solution:

I = \int\limits_{0}^{\pi/2} \; \frac{dx}{1+\cot x}

I = \int\limits_{0}^{\pi/2} \frac{\sin x \, dx}{\sin x + \cos x} \left[ \cot x = \frac{\cos x}{\sin x} \right]

.....(i)

∵

\int\limits_{a}^{b} f(x) \, dx = \int\limits_{a}^{b} f(a+b-x) \, dx

Now,

I = \int\limits_{0}^{\pi/2} \frac{\sin(\pi/2 - x) \, dx}{\sin(\pi/2 - x)} + \cos(\pi/2 - x)

I = \int\limits_{0}^{\pi/2} \frac{\cos x \, dx}{\cos x + \sin x}

.........(ii)

On adding eqs. (i) and (ii), we get

2I = \int\limits_{0}^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x} \, dx = \int\limits_{0}^{\pi/2} 1 \, dx

⇒

2I = [x]_{0}^{\pi/2} \Rightarrow 2I = \pi/2 - 0 \Rightarrow 2I = \pi/2 \Rightarrow I = \pi/4

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