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

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

Question
Let In=∫0∞e−xsin⁡nx dxI_{n}=\displaystyle\int_{0}^{\infty}e^{-x}\sin^{n}x\,dx, n∈Nn\in\mathbb{N}, n>1n>1. Then I2008I2006\dfrac{I_{2008}}{I_{2006}} equals
A2007×200620082+1\dfrac{2007\times 2006}{2008^{2}+1}
B2008×200720082+1\dfrac{2008\times 2007}{2008^{2}+1}correct
C2006×200420082−1\dfrac{2006\times 2004}{2008^{2}-1}
D2008×200720082−1\dfrac{2008\times 2007}{2008^{2}-1}
Solution
Step 1: Set up a reduction formula by writing sin⁡nx=sin⁡n−1x⋅sin⁡x\sin^{n}x=\sin^{n-1}x\cdot\sin x and integrating by parts. With u=sin⁡nxu=\sin^{n}x, dv=e−x dxdv=e^{-x}\,dx (so v=−e−xv=-e^{-x}), all boundary terms vanish at 00 and ∞\infty, leaving
In=n∫0∞e−xsin⁡n−1xcos⁡x dx.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−1xcos⁡xu=\sin^{n-1}x\cos x and dv=e−xdxdv=e^{-x}dx. Differentiating the product,
ddx(sin⁡n−1xcos⁡x)=(n−1)sin⁡n−2xcos⁡2x−sin⁡nx.\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⁡2x=1−sin⁡2x\cos^{2}x=1-\sin^{2}x gives (n−1)sin⁡n−2x−nsin⁡nx(n-1)\sin^{n-2}x-n\sin^{n}x, so
In=n[(n−1)In−2−n In].I_{n}=n\Big[(n-1)I_{n-2}-n\,I_{n}\Big].
 Step 3: Collect InI_{n}:
In=n(n−1)In−2−n2In ⇒ (1+n2)In=n(n−1)In−2 ⇒ InIn−2=n(n−1)1+n2.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=2008n=2008:
I2008I2006=2008×200720082+1.\dfrac{I_{2008}}{I_{2006}}=\dfrac{2008\times 2007}{2008^{2}+1}.
 Correct answer: (2)
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