Indefinite IntegrationhardFree

∫(sin2x+sin4x−sin6x)/(1+cos2x+cos4x+cos6x) | JEE

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

Question
sin2x+sin4xsin6x1+cos2x+cos4x+cos6xdx\displaystyle\int \dfrac{\sin 2x+\sin 4x-\sin 6x}{1+\cos 2x+\cos 4x+\cos 6x}\,dx equals
A13lnsec3x+12lnsec2x+lnsecx+C\dfrac{1}{3}\ln|\sec 3x|+\dfrac{1}{2}\ln|\sec 2x|+\ln|\sec x|+C
B13lnsec3x12lnsec2xlnsecx+C\dfrac{1}{3}\ln|\sec 3x|-\dfrac{1}{2}\ln|\sec 2x|-\ln|\sec x|+Ccorrect
CC13lnsec3x+12lnsec2x+lnsecxC-\dfrac{1}{3}\ln|\sec 3x|+\dfrac{1}{2}\ln|\sec 2x|+\ln|\sec x|
DC13lnsec3x12lnsec2x+lnsecxC-\dfrac{1}{3}\ln|\sec 3x|-\dfrac{1}{2}\ln|\sec 2x|+\ln|\sec x|
Solution
Step 1: Factor the numerator. Pair sin2xsin6x=2cos4xsin2x\sin 2x-\sin 6x=-2\cos 4x\sin 2x and keep sin4x=2sin2xcos2x\sin 4x=2\sin 2x\cos 2x:
Num=2sin2x(cos2xcos4x)=2sin2x(2sin3xsinx)=4sinxsin2xsin3x\text{Num}=2\sin 2x(\cos 2x-\cos 4x)=2\sin 2x\,(2\sin 3x\sin x)=4\sin x\sin 2x\sin 3x
Step 2: Factor the denominator. Pair 1+cos6x=2cos23x1+\cos 6x=2\cos^{2}3x and cos2x+cos4x=2cos3xcosx\cos 2x+\cos 4x=2\cos 3x\cos x:
Den=2cos23x+2cos3xcosx=2cos3x(cos3x+cosx)\text{Den}=2\cos^{2}3x+2\cos 3x\cos x=2\cos 3x(\cos 3x+\cos x)
=2cos3x(2cos2xcosx)=4cosxcos2xcos3x=2\cos 3x\,(2\cos 2x\cos x)=4\cos x\cos 2x\cos 3x
Step 3: Form the ratio:
NumDen=4sinxsin2xsin3x4cosxcos2xcos3x=tanxtan2xtan3x\dfrac{\text{Num}}{\text{Den}}=\dfrac{4\sin x\sin 2x\sin 3x}{4\cos x\cos 2x\cos 3x}=\tan x\tan 2x\tan 3x
Step 4: Use 3x=2x+x3x=2x+x, so tan3x(1tanxtan2x)=tanx+tan2x\tan 3x(1-\tan x\tan 2x)=\tan x+\tan 2x, giving
tanxtan2xtan3x=tan3xtan2xtanx\tan x\tan 2x\tan 3x=\tan 3x-\tan 2x-\tan x
Step 5: Integrate each term with tan(ax)dx=1alnsecax\displaystyle\int\tan(ax)\,dx=\dfrac{1}{a}\ln|\sec ax|:
I=(tan3xtan2xtanx)dx=13lnsec3x12lnsec2xlnsecx+CI=\int(\tan 3x-\tan 2x-\tan x)\,dx=\dfrac{1}{3}\ln|\sec 3x|-\dfrac{1}{2}\ln|\sec 2x|-\ln|\sec x|+C
Correct answer: (2)
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