How do you integrate ∫cot2xdx?
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.