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∫ (1+sinx)/(cosx√cos2x) dx, −π/4 to π/4 | JEE

JEE Maths question with a full step-by-step solution.

Question

The value of

\displaystyle\int_{-\pi/4}^{\pi/4}\dfrac{1+\sin x}{\cos x\,\sqrt{\cos 2x}}\,dx

0

\dfrac{\pi}{4}

\dfrac{\pi}{2}

\pi

correct

Solution

Step 1: Separate the integrand into even and odd parts. Over the symmetric interval

\left[-\dfrac{\pi}{4},\dfrac{\pi}{4}\right]

\dfrac{1+\sin x}{\cos x\sqrt{\cos 2x}}=\underbrace{\dfrac{1}{\cos x\sqrt{\cos 2x}}}_{\text{even}}+\underbrace{\dfrac{\sin x}{\cos x\sqrt{\cos 2x}}}_{\text{odd}}.

The odd part integrates to

0

, so

I=2\int_{0}^{\pi/4}\dfrac{dx}{\cos x\sqrt{\cos 2x}}.

Step 2: Express

\cos 2x

through

\tan x

. Using

\cos 2x=\cos^{2}x\,(1-\tan^{2}x)

, for

0\le x<\dfrac{\pi}{4}

where

\cos x>0

\sqrt{\cos 2x}=\cos x\,\sqrt{1-\tan^{2}x}\ \Rightarrow\ \dfrac{1}{\cos x\sqrt{\cos 2x}}=\dfrac{1}{\cos^{2}x\,\sqrt{1-\tan^{2}x}}=\dfrac{\sec^{2}x}{\sqrt{1-\tan^{2}x}}.

Step 3: Substitute

t=\tan x

dt=\sec^{2}x\,dx

, with

t:0\to 1

I=2\int_{0}^{1}\dfrac{dt}{\sqrt{1-t^{2}}}.

Step 4: Evaluate:

I=2\big[\sin^{-1}t\big]_{0}^{1}=2\left(\dfrac{\pi}{2}-0\right)=\pi.

Correct answer: (4)

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