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If y(x)=int(pi^2/16)^(x^2) (cosx cos sqr...

If `y(x)=int_(pi^2/16)^(x^2) (cosx cos sqrt(theta))/(1+sin^2 sqrt(theta)) d theta` then find `y'(pi)`

Text Solution

Verified by Experts

Differentiating both sides w.r.t. `x`,
`y'(x)=-sinx int_(pi^(2)//16)^(x^(2))(cossqrt(theta))/(1+sin^(2)sqrt(theta))d theta+(cosx.cosx)/(1+sin^(2)x)2x-0`
`y'(pi)=0+((-1)(-1))/(1+0)(2pi)`
`=2pi`
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