Evaluate `int 1/(sin^2theta.cos^2theta) d theta`

int (sin theta + cos theta) * d theta

Evaluate int((2 sin theta + sin 2 theta)d theta)/((cos theta-1)sqrt(cos theta + cos^(2) theta + cos^(3) theta)) .

Evaluate: int cos2 theta ln((cos theta+sin theta)/(cos theta-sin theta))d theta

int(sin^(6)theta+cos^(6)theta)/(sin^(2)theta cos^(2)theta)d theta=

I=int_(0)^( pi/4)(sin2 theta)/(sin^(2)theta+cos^(4)theta)d theta

Find :int(cos theta)/((4+sin^(2)theta)(5-4cos^(2)theta))d theta

int(2sin theta cos theta)/(sin^(4)theta+cos^(4)theta)d theta

int _(0) ^(pi) sin ^(2) theta cos theta d theta

int (sin theta cos theta) / (cos ^ (2) theta-cos theta-2) d theta

Let the be an integrable function defined on [0,a] if I_(1)=int_(pi//2)^(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