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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

g(x)=\left(4\cos^{4}x-2\cos 2x-\dfrac{1}{2}\cos 4x-x^{7}\right)^{1/7}

, then

\displaystyle\int g(g(x))\,dx

2x+C

\dfrac{x^{2}}{2}+C

correct

-2\cos x+C

2\sin x+C

Solution

Step 1: Reduce

4\cos^{4}x

to multiple angles. Using

2\cos^{2}x=1+\cos 2x

4\cos^{4}x=(2\cos^{2}x)^{2}=(1+\cos 2x)^{2}=1+2\cos 2x+\cos^{2}2x

Replace

\cos^{2}2x=\dfrac{1+\cos 4x}{2}

4\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:

4\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}

\Rightarrow g(x)=\left(\dfrac{3}{2}-x^{7}\right)^{1/7}

Step 3: Compose

g

with itself. The outer

g

replaces "

x^{7}

" by

[g(x)]^{7}=\dfrac{3}{2}-x^{7}

g(g(x))=\left(\dfrac{3}{2}-\Big(\dfrac{3}{2}-x^{7}\Big)\right)^{1/7}=(x^{7})^{1/7}=x

Step 4: Integrate:

\int g(g(x))\,dx=\int x\,dx=\dfrac{x^{2}}{2}+C

Correct answer: (2)

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