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Question

How do you integrate cot2xdx?


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Solution

Step- 1: Simplify the expression:

The given integration is,

cot2xdx=cosec2x-1dx[1+cot2x=cosec2x]

=cosec2xdx-dx

=1sin2xdx-x

=cos2xsin2x.cos2xdx-x

=sec2xtan2xdx-x[tanx=sinxcosx,secx=1cosx]

Let u=tanx.

du=sec2xdx[ddxtanx=sec2x]

Step- 2: Evaluate the integration:

Substituting this in the above equation we get,

I=1u2du-x

=u-2+1-2+1-x+c [xndx=xn+1n+1,cbetheintegrationconstant]

=-1u-x+c

=-1tanx-x+c [u=tanx]

=-cotx-x+c

Hence, the required answer is -cotx-x+c.


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