Definite IntegrationhardFree

√(I₂₀₀₁:I₂₀₀₂) for Iₙ=∫cosⁿx cos(nx)dx | JEE

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

Question

Let

I_{n}=\displaystyle\int_{0}^{\pi/2}\cos^{n}x\,\cos(nx)\,dx

n\in\mathbb{N}

. Then

\sqrt{I_{2001}:I_{2002}}

can be the eccentricity of

Aa parabola

Ban ellipse

Ca circle

Da hyperbolacorrect

Solution

Step 1: Build a reduction formula. Write

\cos\!\big((n+1)x\big)=\cos(nx)\cos x-\sin(nx)\sin x

, so

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

Integrating the second integral by parts (with

u=\sin(nx)

dv=\cos^{n+1}x\sin x\,dx

) and simplifying leads to the clean relation

I_{n+1}=\dfrac12\,I_{n}.

Step 2: Verify the base value and solve the recurrence. Since

I_{0}=\displaystyle\int_{0}^{\pi/2}1\,dx=\dfrac{\pi}{2}

, the recurrence

I_{n+1}=\dfrac12 I_{n}

gives

I_{n}=\dfrac{\pi}{2^{\,n+1}}.

(Check:

I_{1}=\displaystyle\int_{0}^{\pi/2}\cos^{2}x\,dx=\dfrac{\pi}{4}

and

I_{2}=\dfrac{\pi}{8}

, both matching.) Step 3: Form the required ratio:

I_{2001}:I_{2002}=\dfrac{\pi}{2^{2002}}:\dfrac{\pi}{2^{2003}}=2^{2003}:2^{2002}=2:1.

Step 4: Take the square root:

\sqrt{I_{2001}:I_{2002}}=\sqrt2.

An eccentricity of

\sqrt2>1

corresponds to a hyperbola. Correct answer: (4)

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