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`u=int_0^(pi/2)cos((2pi)/3sin^2x)dx` and `v=int_0^(pi/2) cos(pi/3 sinx) dx`A. 2u = vB. 2u = 3vC. u = yD. u = 2v |
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Answer» Correct Answer - 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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