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Question

I=sinxcosx1sin4xdx is equal to

A
I=14log(sec2x+tan2x)+C
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B
I=14log(sec2xtan2x)+C
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C
I=12log(sec2x+tan2x)+C
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D
I=12log(sec2xtan2x)+C
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Solution

The correct option is B I=14log(sec2x+tan2x)+C

Consider the following Equation.

I=sinxcosx1sin4xdx


Let t=sin2x

12sinxcosxdt=dx


Therefore,

=sinxcosx1sin4x×12sinxcosxdt

=1211t2dt

=12[12log(1+t1t)]+C

=14log(1+t1t)+C

On putting the value of t, we get

=14log(1+sin2x1sin2x)+C

=14log(1+sin2xcos2x)+C

=14log(sec2x+tan2x)+C


Hence, this is the correct answer.


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