InterviewSolution
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InterviewSolution
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`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` |
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Answer» Correct Answer - 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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