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Find A in ∫(1+cos8x)/(cot2x−tan2x)dx | JEE

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Question
If ∫1+cos⁡8xcot⁡2x−tan⁡2x dx=Acos⁡(8x)+k\displaystyle\int \dfrac{1+\cos 8x}{\cot 2x-\tan 2x}\,dx=A\cos(8x)+k (kk arbitrary), then AA equals
A18\dfrac{1}{8}
B116\dfrac{1}{16}
C−116-\dfrac{1}{16}correct
D−18-\dfrac{1}{8}
Solution
Step 1: Differentiate both sides of ∫1+cos⁡8xcot⁡2x−tan⁡2x dx=Acos⁡8x+k\displaystyle\int\dfrac{1+\cos 8x}{\cot 2x-\tan 2x}\,dx=A\cos 8x+k:
1+cos⁡8xcot⁡2x−tan⁡2x=ddx(Acos⁡8x)=−8Asin⁡8x\dfrac{1+\cos 8x}{\cot 2x-\tan 2x}=\dfrac{d}{dx}\big(A\cos 8x\big)=-8A\sin 8x
 Step 2: Simplify the denominator:
cot⁡2x−tan⁡2x=cos⁡2xsin⁡2x−sin⁡2xcos⁡2x=cos⁡22x−sin⁡22xsin⁡2xcos⁡2x=cos⁡4x12sin⁡4x=2cos⁡4xsin⁡4x\cot 2x-\tan 2x=\dfrac{\cos 2x}{\sin 2x}-\dfrac{\sin 2x}{\cos 2x}=\dfrac{\cos^{2}2x-\sin^{2}2x}{\sin 2x\cos 2x}=\dfrac{\cos 4x}{\tfrac{1}{2}\sin 4x}=\dfrac{2\cos 4x}{\sin 4x}
 Step 3: Simplify the numerator 1+cos⁡8x=2cos⁡24x1+\cos 8x=2\cos^{2}4x, then form the left side:
2cos⁡24x2cos⁡4xsin⁡4x=cos⁡4xsin⁡4x=12sin⁡8x\dfrac{2\cos^{2}4x}{\tfrac{2\cos 4x}{\sin 4x}}=\cos 4x\sin 4x=\dfrac{1}{2}\sin 8x
 Step 4: Equate to the right side:
12sin⁡8x=−8Asin⁡8x  ⇒  −8A=12  ⇒  A=−116\dfrac{1}{2}\sin 8x=-8A\sin 8x\;\Rightarrow\;-8A=\dfrac{1}{2}\;\Rightarrow\; A=-\dfrac{1}{16}
 Correct answer: (3)
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