int cos 2 theta.ln((costheta+sintheta)/(...

`int cos 2 theta.ln((costheta+sintheta)/(costheta-sintheta)) d theta`

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The correct Answer is:

`|cos 2 theta|+c`

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int (cos2theta) . log((costheta+sintheta)/(costheta-sintheta))d theta is equal to

`(costheta-sintheta)^(2)log((costheta+sintheta)/(costheta-sintheta))+C`

`(costheta+sintheta)^(2)log((costheta+sintheta)/(costheta-sintheta))+C`

`((costheta-sintheta)^(2))/(2)log((costheta-sintheta)/(costheta+sintheta))+C`

`(1)/(2)sin2thetalogtan((pi)/(4)+theta)-(1)/(2)logsec2 theta +C`

`(costheta-sintheta)^2log((costheta+sintheta)/(costheta-sintheta))+C`

`(costheta+sintheta)^2log((costheta+sintheta)/(costheta-sintheta))+C`

`((costheta-sintheta)^2)/2log""((costheta-sintheta)/(costheta+sintheta))+C`

`1/2sin2thetalogtan(pi/4+theta)-1/2logsec2theta+C`

`(1)/(2)tantheta`

`(1)/(2)cottheta`

`tantheta`

`cottheta`