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Integral of g(g(x)) with a Self-Inverse | JEE

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

Question
If g(x)=(4cos4x2cos2x12cos4xx7)1/7g(x)=\left(4\cos^{4}x-2\cos 2x-\dfrac{1}{2}\cos 4x-x^{7}\right)^{1/7}, then g(g(x))dx\displaystyle\int g(g(x))\,dx is
A2x+C2x+C
Bx22+C\dfrac{x^{2}}{2}+Ccorrect
C2cosx+C-2\cos x+C
D2sinx+C2\sin x+C
Solution
Step 1: Reduce 4cos4x4\cos^{4}x to multiple angles. Using 2cos2x=1+cos2x2\cos^{2}x=1+\cos 2x,
4cos4x=(2cos2x)2=(1+cos2x)2=1+2cos2x+cos22x4\cos^{4}x=(2\cos^{2}x)^{2}=(1+\cos 2x)^{2}=1+2\cos 2x+\cos^{2}2x
Replace cos22x=1+cos4x2\cos^{2}2x=\dfrac{1+\cos 4x}{2}:
4cos4x=1+2cos2x+12+cos4x2=32+2cos2x+cos4x24\cos^{4}x=1+2\cos 2x+\dfrac{1}{2}+\dfrac{\cos 4x}{2}=\dfrac{3}{2}+2\cos 2x+\dfrac{\cos 4x}{2}
Step 2: Subtract the remaining trigonometric terms:
4cos4x2cos2x12cos4x=(32+2cos2x+cos4x2)2cos2xcos4x2=324\cos^{4}x-2\cos 2x-\dfrac{1}{2}\cos 4x=\left(\dfrac{3}{2}+2\cos 2x+\dfrac{\cos 4x}{2}\right)-2\cos 2x-\dfrac{\cos 4x}{2}=\dfrac{3}{2}
g(x)=(32x7)1/7\Rightarrow g(x)=\left(\dfrac{3}{2}-x^{7}\right)^{1/7}
Step 3: Compose gg with itself. The outer gg replaces "x7x^{7}" by [g(x)]7=32x7[g(x)]^{7}=\dfrac{3}{2}-x^{7}:
g(g(x))=(32(32x7))1/7=(x7)1/7=xg(g(x))=\left(\dfrac{3}{2}-\Big(\dfrac{3}{2}-x^{7}\Big)\right)^{1/7}=(x^{7})^{1/7}=x
Step 4: Integrate:
g(g(x))dx=xdx=x22+C\int g(g(x))\,dx=\int x\,dx=\dfrac{x^{2}}{2}+C
Correct answer: (2)
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