Definite IntegrationhardFree

Ratio I₂₀₀₈/I₂₀₀₆ for ∫e⁻ˣsinⁿx dx | JEE

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

Question

Let

I_{n}=\displaystyle\int_{0}^{\infty}e^{-x}\sin^{n}x\,dx

n\in\mathbb{N}

n>1

. Then

\dfrac{I_{2008}}{I_{2006}}

equals

\dfrac{2007\times 2006}{2008^{2}+1}

\dfrac{2008\times 2007}{2008^{2}+1}

correct

\dfrac{2006\times 2004}{2008^{2}-1}

\dfrac{2008\times 2007}{2008^{2}-1}

Solution

Step 1: Set up a reduction formula by writing

\sin^{n}x=\sin^{n-1}x\cdot\sin x

and integrating by parts. With

u=\sin^{n}x

dv=e^{-x}\,dx

(so

v=-e^{-x}

), all boundary terms vanish at

0

and

\infty

, leaving

I_{n}=n\int_{0}^{\infty}e^{-x}\sin^{n-1}x\cos x\,dx.

Step 2: Integrate by parts a second time on the new integral, taking

u=\sin^{n-1}x\cos x

and

dv=e^{-x}dx

. Differentiating the product,

\dfrac{d}{dx}\big(\sin^{n-1}x\cos x\big)=(n-1)\sin^{n-2}x\cos^{2}x-\sin^{n}x.

Replacing

\cos^{2}x=1-\sin^{2}x

gives

(n-1)\sin^{n-2}x-n\sin^{n}x

, so

I_{n}=n\Big[(n-1)I_{n-2}-n\,I_{n}\Big].

Step 3: Collect

I_{n}

I_{n}=n(n-1)I_{n-2}-n^{2}I_{n}\ \Rightarrow\ \big(1+n^{2}\big)I_{n}=n(n-1)I_{n-2}\ \Rightarrow\ \dfrac{I_{n}}{I_{n-2}}=\dfrac{n(n-1)}{1+n^{2}}.

Step 4: Put

n=2008

\dfrac{I_{2008}}{I_{2006}}=\dfrac{2008\times 2007}{2008^{2}+1}.

Correct answer: (2)

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