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u=int0^(pi/2)cos((2pi)/3sin^2x)dx and v=...

`u=int_0^(pi/2)cos((2pi)/3sin^2x)dx` and `v=int_0^(pi/2) cos(pi/3 sinx) dx`

A

2u = v

B

2u = 3v

C

u = y

D

u = 2v

Text Solution

Verified by Experts

The correct Answer is:
A
`u=int_(0)^(pi//2)cos((2pi)/(3)sin^(2)x)dx`
`u=int_(0)^(pi//2)cos((2pi)/(3)cos^(2)x)dx`
`rArr" "2u=int_(0)^(pi//2)[cos((2pi)/(3)sin^(2)x)+cos((2pi)/(3).cos^(2)x)]dx`
`rArr" "2u=int_(0)^(pi//2)2cos.(pi)/(3).cos((pi)/(3)cos2x)dx`
`rArr" "u=(1)/(2)int_(0)^(pi)cos((pi)/(3)cost)dt" [Put 2x = t]"`
`=int_(0)^(pi//2)cos((pi)/(3)cost)dt`
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