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Iff(x)=int((x^2)/(16))^(x^2)(sinxsinsqrt...

`Iff(x)=int_((x^2)/(16))^(x^2)(sinxsinsqrt(theta))/(1+cos^2sqrt(theta))d theta,` then find the value of `f^(prime)(pi/2)dot`

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Verified by Experts

The correct Answer is:
`pi`
`f(x)=sinx int_(x^(2)//16)^(x^(2))(sin sqrt(theta))/(1+cos^(2)sqrt(theta)) d theta`
`:.f'(x)=sinx[(sinx)/(1+cos^(2)x)2x-0]+(int_(pi^(2)//16)^(x^(2))(sinsqrt(theta))/(1+cos^(2)sqrt(theta)) d theta)cosx`
Therefore `f'((pi)/2)=pi`
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