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

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Question
The value of ∫−π/4π/41+sin⁡xcos⁡x cos⁡2x dx\displaystyle\int_{-\pi/4}^{\pi/4}\dfrac{1+\sin x}{\cos x\,\sqrt{\cos 2x}}\,dx is
A00
Bπ4\dfrac{\pi}{4}
Cπ2\dfrac{\pi}{2}
Dπ\picorrect
Solution
Step 1: Separate the integrand into even and odd parts. Over the symmetric interval [−π4,π4]\left[-\dfrac{\pi}{4},\dfrac{\pi}{4}\right],
1+sin⁡xcos⁡xcos⁡2x=1cos⁡xcos⁡2x⏟even+sin⁡xcos⁡xcos⁡2x⏟odd.\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 00, so
I=2∫0π/4dxcos⁡xcos⁡2x.I=2\int_{0}^{\pi/4}\dfrac{dx}{\cos x\sqrt{\cos 2x}}.
 Step 2: Express cos⁡2x\cos 2x through tan⁡x\tan x. Using cos⁡2x=cos⁡2x (1−tan⁡2x)\cos 2x=\cos^{2}x\,(1-\tan^{2}x), for 0≤x<π40\le x<\dfrac{\pi}{4} where cos⁡x>0\cos x>0,
cos⁡2x=cos⁡x 1−tan⁡2x ⇒ 1cos⁡xcos⁡2x=1cos⁡2x 1−tan⁡2x=sec⁡2x1−tan⁡2x.\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⁡xt=\tan x, dt=sec⁡2x dxdt=\sec^{2}x\,dx, with t:0→1t:0\to 1:
I=2∫01dt1−t2.I=2\int_{0}^{1}\dfrac{dt}{\sqrt{1-t^{2}}}.
 Step 4: Evaluate:
I=2[sin⁡−1t]01=2(π2−0)=π.I=2\big[\sin^{-1}t\big]_{0}^{1}=2\left(\dfrac{\pi}{2}-0\right)=\pi.
 Correct answer: (4)
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