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

1 Answer
Apr 13, 2018

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