Definite IntegrationeasyFree

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

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

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)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Definite Integration · easy
∫0π/2sin⁡x1+cos⁡x+sin⁡x dx\displaystyle\int_{0}^{\pi/2}\dfrac{\sin x}{1+\cos x+\sin x}\,dx equals
Definite Integration · easy
If f(x)=∫1/xxcos⁡t2 dtf(x)=\displaystyle\int_{1/x}^{\sqrt{x}}\cos t^{2}\,dt for x>0x>0, then df(x)dx\dfrac{df(x)}{dx} is
Definite Integration · easy
If ∫sin⁡x1t2f(t) dt=1−sin⁡x\displaystyle\int_{\sin x}^{1}t^{2}f(t)\,dt=1-\sin x for all x∈(0,π2]x\in\left(0,\dfrac{\pi}{2}\right], then the value of f ⁣(13)f\!\left(\dfrac{1}{\sqrt3}\right) is
Definite Integration · easy
Let f:(0,∞)→Rf:(0,\infty)\to\mathbb{R} and F(x)=∫0xf(t) dtF(x)=\displaystyle\int_{0}^{x}f(t)\,dt. If F ⁣(x2)=x2(1+x)F\!\big(x^{2}\big)=x^{2}(1+x), then f(4)f(4) equals

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath