How do you find the indefinite integral of #int sintheta/sqrt(1-costheta)#?

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Answer 1 · 2016-12-26T10:23:49

#int (sin theta d theta) /sqrt(1-cos theta) = 2sqrt(1-cos theta)+C#

Note that:

#sintheta d theta = d(-costheta)#

You can then perform the substitution:

#x=-costheta#

and the integral becomes:

#int (sin theta d theta) /sqrt(1-cos theta) =int (dx)/sqrt(1+x)=int (1+x)^(-1/2)d(1+x)=2(1+x)^(1/2)+C#

and substituting back:

#int (sin theta d theta) /sqrt(1-cos theta) = 2sqrt(1-cos theta)+C#