Definite IntegrationeasyFree

Find n: ∫sin¹⁰x cos¹⁰x = 2⁻ⁿ∫(sin¹⁰+cos¹⁰) | JEE

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

Question

\displaystyle\int_{0}^{\pi/2}\sin^{10}x\,\cos^{10}x\,dx=2^{-n}\displaystyle\int_{0}^{\pi/2}\big(\sin^{10}x+\cos^{10}x\big)\,dx

n\in\mathbb{N}

, then

n=

7

9

10

11

correct

Solution

Step 1: Rewrite the left side using

\sin x\cos x=\dfrac{\sin 2x}{2}

\int_{0}^{\pi/2}\sin^{10}x\cos^{10}x\,dx=\int_{0}^{\pi/2}\Big(\dfrac{\sin 2x}{2}\Big)^{10}dx=\dfrac{1}{2^{10}}\int_{0}^{\pi/2}\sin^{10}2x\,dx.

Step 2: Substitute

2x=u

dx=\dfrac{du}{2}

, with

u:0\to\pi

, and use the symmetry

\displaystyle\int_{0}^{\pi}\sin^{10}u\,du=2\int_{0}^{\pi/2}\sin^{10}u\,du

\dfrac{1}{2^{10}}\cdot\dfrac12\int_{0}^{\pi}\sin^{10}u\,du=\dfrac{1}{2^{10}}\int_{0}^{\pi/2}\sin^{10}u\,du.

Step 3: Simplify the right side. By the complementary symmetry

\displaystyle\int_{0}^{\pi/2}\cos^{10}x\,dx=\int_{0}^{\pi/2}\sin^{10}x\,dx

, so

2^{-n}\int_{0}^{\pi/2}\big(\sin^{10}x+\cos^{10}x\big)\,dx=2^{-n}\cdot 2\int_{0}^{\pi/2}\sin^{10}x\,dx=2^{-n+1}\int_{0}^{\pi/2}\sin^{10}x\,dx.

Step 4: Match the two sides. Cancelling the common

\displaystyle\int_{0}^{\pi/2}\sin^{10}x\,dx

2^{-10}=2^{-n+1}\ \Rightarrow\ -10=-n+1\ \Rightarrow\ n=11.

Correct answer: (4)

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