`Ify(x)=int_(pi^2/16)^(x^2)(cosxcossqrt(theta))/(1+sin^2sqrt(theta)),t h e nfin d(dy)/(dx)a tx=pidot`

If y(x)=\ int_(pi^2//16)^(x^2)(cos xdotcossqrt(theta))/(1+sin^2sqrt(theta))dotdtheta\ t h e n\ fin d(dy)/(dx)a t\ x=pi

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

x=sqrt(sin2 theta),y=sqrt(cos2 theta) then find (dy)/(dx)

If f(x)=int_((x^(2))/(16))^(x^(2))(sin x sin sqrt(theta))/(1+cos^(2)sqrt(theta))d theta, then find the value of f'((pi)/(2))^((16)/(16))

int_(pi^(2)//16)^(pi^(2)//4)(sin sqrt(x))/(sqrt(x))dx=

If f(x)=int_(0)^((pi)/(2))(ln(1+x sin^(2)theta))/(sin^(2)theta)d theta,x>=0 then :

int_(0)^(pi//2) (cos theta)/(sqrt(4-sin^(2)theta))d theta is equal to

int_(pi//4)^(pi//2)(costheta)/((cos\ (theta)/(2)+sin\ (theta)/(2)))\ d theta

int_(0)^( pi)(sin theta+cos theta)/(sqrt(1+sin2 theta))d theta

If x=2cos theta -cos 2theta ,y=2sin theta -sin 2theta ,then (dy)/(dx) =