KVPY SX SB Exam 20141 Question Paper

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Question : 87 of 120

Marks: +1, -0

Let

I _n = ∫↖{π\/2}↙{0}x_n \cosx \dx

, where n is a non-negative integer.
Then

∑↖{∞}↙{n=2} (I_n/n!+I_{n-2}/(n-2)!)

equals

$e^{π\/2}-1-π/2$
$e^{π\/2}-1$
$e^{π\/2} -π/2$
$e^{π\/2}$

Solution:

I_n = ∫ ↖{π\/ 2} ↙{0} {x^n}↙{\I}\ {\cosx}↙{ \II} dx

x^n \sin x|^{π\/ 2}_0-∫ ↖{π\/ 2} ↙{0}n x^{n-1} \sin x dx

=(π/2)^2-0-( nx^{n-1}(-\cosx)|^{π\/ 2}_0

-∫ ↖{π\/ 2} ↙{0}n ({n-1})x^{n-2}(-\cosx)dx

=(π/2)^2-0-n(n-1)∫ ↖{π\/ 2} ↙{0}x^{n-2} {\cosx \dx}

I_n=( {π/ 2})^n-n(n -1)I_{n-2}

∑↖{∞}↙{n=2} (I_n/n!+I_{n-2}/(n-2)!)

=∑↖{∞}↙{n=2}({( {π/ 2})^n-n(n -1)I_{n-2}}/n!+I_{n-2}/(n-2)!)

∑↖{∞}↙{n=2}((π/2)^n\ 1/n!)

=( π/2)^2\ 1/2!+(π/2)^3\ 1/3!+(π/2)^4 \ 1/4!+....

=e^{π\/2}-1-π/2

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