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cos(x+π/6)cos2x·cos(x+5π/6) dx | JEE

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

Question
∫4cos⁡ ⁣(x+π6)cos⁡2x cos⁡ ⁣(x+5π6)dx\displaystyle\int 4\cos\!\left(x+\dfrac{\pi}{6}\right)\cos 2x\,\cos\!\left(x+\dfrac{5\pi}{6}\right)dx is
A−(x+sin⁡4x4+sin⁡2x2)+C-\left(x+\dfrac{\sin 4x}{4}+\dfrac{\sin 2x}{2}\right)+Ccorrect
B−(x+sin⁡4x4−sin⁡2x2)+C-\left(x+\dfrac{\sin 4x}{4}-\dfrac{\sin 2x}{2}\right)+C
C−(x−sin⁡4x4+sin⁡2x2)+C-\left(x-\dfrac{\sin 4x}{4}+\dfrac{\sin 2x}{2}\right)+C
D−(x−sin⁡4x4−sin⁡2x2)+C-\left(x-\dfrac{\sin 4x}{4}-\dfrac{\sin 2x}{2}\right)+C
Solution
Step 1: Combine the two outer cosines using 2cos⁡Acos⁡B=cos⁡(A−B)+cos⁡(A+B)2\cos A\cos B=\cos(A-B)+\cos(A+B), with A=x+π6, B=x+5π6A=x+\dfrac{\pi}{6},\ B=x+\dfrac{5\pi}{6}:
2cos⁡ ⁣(x+π6)cos⁡ ⁣(x+5π6)=cos⁡ ⁣(−2π3)+cos⁡(2x+π)=−12−cos⁡2x2\cos\!\left(x+\tfrac{\pi}{6}\right)\cos\!\left(x+\tfrac{5\pi}{6}\right)=\cos\!\left(-\tfrac{2\pi}{3}\right)+\cos(2x+\pi)=-\dfrac{1}{2}-\cos 2x
 Step 2: Reinsert the remaining factors (the leading 4=2⋅24=2\cdot 2 and the middle cos⁡2x\cos 2x):
integrand=2(−12−cos⁡2x)cos⁡2x=−cos⁡2x−2cos⁡22x\text{integrand}=2\left(-\dfrac{1}{2}-\cos 2x\right)\cos 2x=-\cos 2x-2\cos^{2}2x
 Step 3: Replace 2cos⁡22x=1+cos⁡4x2\cos^{2}2x=1+\cos 4x:
=−cos⁡2x−(1+cos⁡4x)=−(1+cos⁡2x+cos⁡4x)=-\cos 2x-(1+\cos 4x)=-(1+\cos 2x+\cos 4x)
 Step 4: Integrate term by term:
I=−∫(1+cos⁡2x+cos⁡4x) dx=−(x+sin⁡2x2+sin⁡4x4)+CI=-\int(1+\cos 2x+\cos 4x)\,dx=-\left(x+\dfrac{\sin 2x}{2}+\dfrac{\sin 4x}{4}\right)+C
 Correct answer: (1)
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