pi)int(d theta)/(b cos theta+c sin theta)du

int (sin theta + cos theta) * d theta

If : sin^(2) ((pi)/(4) + (theta)/(2)) - cos^(2) ((pi)/(4) + (theta)/(2))= A) sin theta B) cos theta C) sin 2 theta D) cos 2 theta

int_(0)^(2pi) (sin 2 theta)/(a-b cos 2 theta )d theta =

Integration using rigonometric identities : int(sin theta)/(cos3 theta)+(sin 3 theta)/(cos9 theta)+( sin9 theta)/(cos 27theta)d theta=...+c

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

int_(pi//4) ^(pi//2) (cos theta)/((cos (theta/2) + sin (theta/2))^(3))d theta=

int cos 2theta log ((cos theta+sin theta)/(cos theta-sin theta))d theta

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

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

int_(0)^((pi)/(2))(cos theta-sin theta)/((1+cos theta)(1+sin theta))d theta equals -