`intcos2thetalog((costheta+sintheta)/(costheta-sintheta))=`

Answer

`(1)/(2)"sin"2 theta "In"(("cos"theta+"sin"theta)/("cos"theta-"sin"theta))+(1)/(2)"In"("cos"2 theta)+c`

Explanation

`I=int cos 2 theta "In"((cos theta+sin theta)/(cos theta-sin theta))d theta" "["given"]`
We integrate it by taking parts In `((cos theta + sin theta)/(cos theta - sin theta))` as first function
` =(sin 2 theta)/(2)"In"((cos theta + sin theta)/(cos theta - sin theta))-(1)/(2)int(d)/(d theta)["In"((cos theta+sin theta)/(cos theta-sin theta))]sin 2 theta d theta" "....(i)`
But `(d)/(d theta)["In"((cos theta+sin theta)/(cos theta-sin theta))]`
`=(d)/(d theta)["In" (cos theta + sin theta)-"In"(cos theta - sin theta)]`
`=(1)/((cos theta+sin theta))*(-sin theta+cos theta)-((-sin theta-cos theta))/(cos theta - sin theta)`
`=((cos theta-sin theta)(cos theta-sin theta)-(cos theta+sin theta)(-sin theta -cos theta))/((cos theta + sin theta)(cos theta - sin theta))`
`=(cos^(2)theta-cos theta sin theta - sin theta cos theta + sin^(2)theta+cos theta sin theta+cos^(2)theta+sin^(2)theta + cos theta * sin theta)/(cos^(2)theta - sin^(2)theta)`
`=(2(cos^(2)theta+sin^(2)theta))/(cos2theta)=(2)/(cos2theta)`
Therefore, on putting this value in Eq.(i), we get `I=(1)/(2)sin 2 theta"In"((cos theta+sin theta)/(cos theta-sin theta))-(1)/(2) int sin 2 theta (2)/(cos2theta)d theta`
`=(1)/(2)sin 2 theta"In"((cos theta+sin theta)/(cos theta-sin theta))+(1)/(2)"In"(cos 2 theta)+c`