Indefinite IntegrationmediumFree

Integral of sinx/(cos²x·√(cos2x)) | JEE

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

Question
∫sin⁡xcos⁡2x cos⁡2x dx\displaystyle\int \dfrac{\sin x}{\cos^{2}x\,\sqrt{\cos 2x}}\,dx equals
AC−1−tan⁡2xC-\sqrt{1-\tan^{2}x}correct
BC−1+sec⁡2xC-\sqrt{1+\sec^{2}x}
CC−1+cos⁡2xC-\sqrt{1+\cos^{2}x}
DC−1+sin⁡2xC-\sqrt{1+\sin^{2}x}
Solution
Step 1: Rewrite the integrand to expose sec⁡x\sec x. Since sin⁡xcos⁡2x=sec⁡xtan⁡x\dfrac{\sin x}{\cos^{2}x}=\sec x\tan x and cos⁡2x=2cos⁡2x−1\cos 2x=2\cos^{2}x-1,
I=∫sec⁡xtan⁡x2cos⁡2x−1 dxI=\int\dfrac{\sec x\tan x}{\sqrt{2\cos^{2}x-1}}\,dx
 Step 2: Substitute t=sec⁡x⇒dt=sec⁡xtan⁡x dxt=\sec x\Rightarrow dt=\sec x\tan x\,dx. Also cos⁡2x=1t2\cos^{2}x=\dfrac{1}{t^{2}}, so
2cos⁡2x−1=2t2−1=2−t2t2  ⇒  2cos⁡2x−1=2−t2∣t∣2\cos^{2}x-1=\dfrac{2}{t^{2}}-1=\dfrac{2-t^{2}}{t^{2}}\;\Rightarrow\;\sqrt{2\cos^{2}x-1}=\dfrac{\sqrt{2-t^{2}}}{|t|}
 Step 3: Assemble:
I=∫dt2−t2/t=∫t dt2−t2I=\int\dfrac{dt}{\sqrt{2-t^{2}}/t}=\int\dfrac{t\,dt}{\sqrt{2-t^{2}}}
 Step 4: Substitute w=2−t2⇒dw=−2t dt⇒t dt=−dw2w=2-t^{2}\Rightarrow dw=-2t\,dt\Rightarrow t\,dt=-\dfrac{dw}{2}:
I=∫−dw/2w=−w+C=−2−t2+CI=\int\dfrac{-dw/2}{\sqrt{w}}=-\sqrt{w}+C=-\sqrt{2-t^{2}}+C
 Step 5: Back-substitute t=sec⁡xt=\sec x; since 2−sec⁡2x=1−tan⁡2x2-\sec^{2}x=1-\tan^{2}x,
I=−2−sec⁡2x+C=C−1−tan⁡2xI=-\sqrt{2-\sec^{2}x}+C=C-\sqrt{1-\tan^{2}x}
 Correct answer: (1)
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