Question

How to evaluate : `int (1+sinx)/[cosx (1+cosx)]` ?

Answer 1

`I=-cscx+ln|secx+tanx|+ln|secx|+ln|sinx|+cotx-ln|cscx-cotx|+c`

Explanation:

Here,

`I=int(1+sinx)/(cosx(1+cosx))dx`

Now,

`M=(1+sinx)/(cosx(1+cosx))...to(say)`

`M=(1+sinx)/(cosx(1+cosx))xx(1-cosx)/(1-cosx)`

`=(1+sinx-cosx-sinxcosx)/(cosx(1-cos^2x))`

#=1/(cosxsin^2x)+sinx/(cosxsin^2x)-cosx/(cosxsin^2x)- (sinxcosx)/(cosxsin^2x)#

`=(cos^2x+sin^2x)/(cosxsin^2x)+1/(sinxcosx)-1/sin^2x-1/sinx`

#=cosx/sin^2x+1/cosx+(sin^2x+cos^2x)/(sinxcosx)-1/sin^2x- 1/sinx#

`=1/sinxcosx/sinx+1/cosx+sinx/cosx+cosx/sinx-1/sin^2x-1/sinx`

`M=cscxcotx+secx+tanx+cotx-csc^2x-cscx`

`I=int(cscxcotx+secx+tanx+cotx-csc^2x-cscx)dx`

`=-cscx+ln|secx+tanx|+ln|secx|+ln|sinx|+cotx-ln|cscx-cotx|+c`