How do you find the integral of #f(x)=1/(1-sinx)# using integration by parts?

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Answer 1 · 2018-04-10T22:49:05

#tanx+secx+C#

#intdx/(1-sinx)#

Let's first get this into a more workable form by multiplying through by the conjugate:

#=int1/(1-sinx)*(1+sinx)/(1+sinx)dx#

#=int(1+sinx)/(1-sin^2x)dx#

Recall that #sin^2x+cos^2x=1#:

#=int(1+sinx)/cos^2xdx#

Splitting up the integral by addition:

#=int1/cos^2xdx+intsinx/cos^2xdx#

Rewrite using the identities #secx=1/cosx# and #tanx=sinx/cosx#:

#=intsec^2xdx+intsecxtanxdx#

And then, from the knowledge that #d/dxtanx=sec^2x# and #d/dxsecx=secxtanx#, we see that:

#=tanx+secx+C#