`Iflambda=int_0^(pi/2)costhetaf(sintheta+cos^2theta)dtheta` `a n dI_2=int_0^(pi//2)sin2thetaf(sintheta+cos^2theta)dtheta,t h e n` `I_1=-2I_2` (b) `I_1=I_2` `2I_1=I_2` (d) `I_1=-I_2`

Evaluate int_(0)^(pi//2)(costheta)/((1+sintheta)(2+sintheta))d theta

If int_0^(pi/2)logsinthetad(theta)=k , then find the value of int_0^(pi/2)(theta/(sintheta))^2d(theta) in terms of k

Evaluate: int_(0)^(pi/2)(costheta)/((1+sintheta)(2+sintheta))d theta .

Let f beintegrable over [0,a] for any real value of a . If I_(1)=int_(0)^(pi//2)cos theta f(sin theta +cos^(2) theta) d theta and I_(2)=int_(0)^(pi//2) sin 2 theta f(sin theta+cos^(2) theta) d theta , then

If int_0^(pi/2)logsinthetadtheta=k , then find the value of int_pi^(pi/2)(theta/(s intheta))^2dtheta in terms of k

Evaluate: int_(-pi/2)^(pi/2)log((a-s intheta)/(a+sintheta))dtheta,a >0

Evaluate: int_0^(pi/4)(tan^(-1)((2cos^2theta)/(2-sin2theta)))sec^2theta d theta dot

I_1=int_0^(pi/2)ln(sinx)dx ,I_2=int_(-pi/4)^(pi/4)ln(sinx+cosx)dxdot Then (a) I_1=2I_2 (b) I_2=2I_1 (c) I_1=4I_2 (d) I_2=4I_1

L e tf(n)=1+1/2+1/3++1/ndot Then show that f(n)=int_0^(pi/2)cot(theta/2)(1-cos^ntheta)d theta