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intcos2thetalog((costheta+sintheta)/(cos...

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

A

`(cos theta-sintheta)^(2)log((cos theta+sin theta)/(cos theta-cos theta))+C`

B

`(cos theta+sin theta)^(2)log((cos theta+sin theta)/(cos theta-sin theta))+C`

C

`((cos theta-sin theta)^(2))/(2)log((cos theta-sin theta)/(cos theta+sin theta))+C`

D

`(1)/(2)sin 2 theta log tan ((pi)/(4+theta)-(1)/(2)log sec 2 theta +C`

Text Solution

Verified by Experts

The correct Answer is:
D
Since, `log((cos theta + sin theta)/(cos theta - sin theta))=log tan ((pi)/(4)+theta)`
`"and "int sec theta d theta = log tan((pi)/(4)+(theta)/(2))`
`rArr" "int sec 2 theta d theta =(1)/(2)log tan ((pi)/(4)+theta)`
`" "2sec 2 theta=(d)/(d theta)log tan ((pi)/(4)+theta)`
By integration by parts
`therefore" "l=(1)/(2)sin 2 theta log tan ((pi)/(4)+theta)-int tan 2 theta d theta`
`=(1)/(2)sin 2 theta log tan ((pi)/(4)+theta)-(1)/(2)log sec 2 theta +C`
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