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inte^(sin theta)[log sin theta+"cosec"^(...

`inte^(sin theta)[log sin theta+"cosec"^(2)theta]cos theta d theta` is equal to

A

`inte^(sin theta)[log sin theta+"cosec"^(2)theta]+C`

B

`e^(sin theta)[log sin theta+"cosec "theta]+C`

C

`e^(sin theta)[log sin theta-"cosec "theta]+C`

D

`e^(sin theta)[log sin theta - "cosec"^(2) theta]+C`

Text Solution

Verified by Experts

The correct Answer is:
C
Let `l=inte^(sin theta)(log sin theta)cos theta d theta+int e^(sin theta)"cosec"^(2)thetacos theta d theta`
Put `sin theta = t rArr cos theta d theta = dt`
`therefore" "l=int e^(t) log t dt +int e^(t)t^(-2)dt`
`=logt e^(t)-int(e^(t))/(t)dt+(e^(t)t^(-1))/(-1)-int(e^(t)t^(-))/(-1)dt`
`=e^(t)(logt-(1)/(t))+c`
`=e^(sin theta)(log sin theta - "cosec" theta )+C`
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