" 8."int cos2 theta*ln(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 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

int cos2 theta log((cos theta+sin theta)/(cos theta-sin theta))=

cos theta+sin theta=cos2 theta+sin2 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=

int(sqrt(cos2 theta))/(sin theta)d theta=

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

int(2sin2 theta-cos theta)/(6-cos^(2)theta-4sin theta)d theta