The value {\int^{\pi/2}_{0}(\sin x)^{\sqrt2+1}dx}/{\int^{\pi/2}_{0}(\sin x)^{\sqrt2-1}dx} is

KVPY SX SB Exam 2012 Question Paper

Question : 12 of 120

Marks: +1, -0

The value

{∫↖{π/2}↙{0}(\sin x)^{√2+1}dx}/{∫↖{π/2}↙{0}(\sin x)^{√2-1}dx}

Solution:

I_1=∫↖{π/2}↙{0}(\sin x)^{√2}.\sin xdx

I_2=∫↖{π/2}↙{0}(\sin x)^{√2-1}dx

I_1=((\sin x)^{√2}∫\sin xdx)^{π\/2}_{0}

-∫↖{π\/2}↙{0}(√2(\sin x)^{√2-1}\cos x∫\sin xdx)

=-(\cos x(\sin x)^{√2})^{π\/2}_{0}

+√2∫↖{π\/2}↙{0}(\sin x)^{√2-1}(1-\sin^2x)dx

I_1/I_2=√2/{1+√2}×{√2-1}/{√2-1}

=2-√2

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