cos2 theta*log((cos theta+sin theta)/(cos theta-sin theta))

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

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

Evaluate: intcos 2 theta ln((cos theta+sin theta)/(cos theta-sin theta))d theta

Evaluate: intcos 2 theta ln((cos theta+sin theta)/(cos theta-sin theta))d theta

cos theta+sin theta=cos2 theta+sin2 theta

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

If cos theta>sin theta>0 then I=int[ln[(1+sin2 theta)/(1-sin2 theta)]^(cos^(2)theta)+ln[(cos2 theta)/(1+sin2 theta)]]K sin2 theta ln[(cos theta+sin theta)/(cos theta-sin theta)]+m ln|cos2 theta|+c then k+m=

Prove (cos theta)/(1-sin theta)=(1+cos theta+sin theta)/(1+cos theta-sin theta)

(cos theta)/(1-sin theta)=(1+cos theta+sin theta)/(1+cos theta-sin theta)

(cos theta + sin theta )/( cos theta - sin theta )-(cos theta - sin theta )/( cos theta + sin theta ) =