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

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Question
If ∫0π/2sin⁡10x cos⁡10x dx=2−n∫0π/2(sin⁡10x+cos⁡10x) dx\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∈Nn\in\mathbb{N}, then n=n=
A77
B99
C1010
D1111correct
Solution
Step 1: Rewrite the left side using sin⁡xcos⁡x=sin⁡2x2\sin x\cos x=\dfrac{\sin 2x}{2}:
∫0π/2sin⁡10xcos⁡10x dx=∫0π/2(sin⁡2x2)10dx=1210∫0π/2sin⁡102x dx.\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=u2x=u, dx=du2dx=\dfrac{du}{2}, with u:0→πu:0\to\pi, and use the symmetry ∫0πsin⁡10u du=2∫0π/2sin⁡10u du\displaystyle\int_{0}^{\pi}\sin^{10}u\,du=2\int_{0}^{\pi/2}\sin^{10}u\,du:
1210⋅12∫0πsin⁡10u du=1210∫0π/2sin⁡10u 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 ∫0π/2cos⁡10x dx=∫0π/2sin⁡10x dx\displaystyle\int_{0}^{\pi/2}\cos^{10}x\,dx=\int_{0}^{\pi/2}\sin^{10}x\,dx, so
2−n∫0π/2(sin⁡10x+cos⁡10x) dx=2−n⋅2∫0π/2sin⁡10x dx=2−n+1∫0π/2sin⁡10x dx.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 ∫0π/2sin⁡10x dx\displaystyle\int_{0}^{\pi/2}\sin^{10}x\,dx,
2−10=2−n+1 ⇒ −10=−n+1 ⇒ n=11.2^{-10}=2^{-n+1}\ \Rightarrow\ -10=-n+1\ \Rightarrow\ n=11.
 Correct answer: (4)
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