If Iₙ = _{π/2}^{∞} e^{-x} ⁿ x dx , then {I₂018}}{I₂016}} =

Correct Answer is : 2018×2017(2018)2+12018× 2017/(2018)²+1(2018)2+12018×2017

Correct Answer is : 2018×2017(2018)2+12018× 2017/(2018)²+1(2018)2+12018×2017

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TS EAMCET 5 May 2018 Shift 1 Solved Paper

Section: Physics

Question : 116 of 160

Marks: +1, -0

I_n

\int\limits_{\frac{\pi}{2}}^{\infty} e^{-x} \cos^{n} x

\frac{I_{2018}}{I_{2016}}

Solution:

The integration is expressed as,

I_n

\int\limits_{\frac{\pi}{2}}^{\infty} e^{-x} \cos^{n} x

Integrate the above integral by parts,

I_n = \left[ -e^{-x} \cos^{n} x \right]_{\frac{\pi}{2}}^{\infty}

-\int\limits_{\frac{\pi}{2}}^{\infty} -e^{-x} \cdot n \cos^{n-1} x (-\sin x) \, dx

=0-\int\limits_{\frac{\pi}{2}}^{\infty} n e^{-x} \cos^{n-1} x \sin x \, dx

Again apply the integration by parts in the above integral,

I_{n}=-n^{2} I_{n}+n(n-1) I_{n-2}

I_{n} (n^{2}+1) = n(n-1) I_{n-2}

\frac{I_n}{I_{n-2}} = \frac{n(n-1)}{n^{2}+1}

Substitute n = 2018 in the above equation,

\frac{I_{2018}}{I_{2016}} = \frac{2018(2017)}{2018^{2}+1}

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