Question

How do you integrate `(sin x)/(1+sin x)^2`?

Answer 1

This is one of the trickiest integrals I've seen in a long time. I think I got a good start on it based on knowing the answer from Wolfram Alpha. Can anybody help me finish it?

Wolfram Alpha (https://www.wolframalpha.com/ ) produces:

`\int (sin(x))/((1+sin(x))^2)\ dx=\frac{3sin(x/2)+3cos(x/2)-2cos((3x)/2)}{3(sin(x/2)+cos(x/2))^3}+C`

Based on this answer, it seems good in the denominator of the original integrand to write (using the double-angle formula for sine after writing `x=2\cdot x/2` and then using the Pythagorean Identity)

`(1+sin(x))^2=(1+2cos(x/2)sin(x/2))^2`

`=(cos^{2}(x/2)+2cos(x/2)sin(x/2)+sin^{2}(x/2))^2`

`=((cos(x/2)+sin(x/2))^2)^2=(cos(x/2)+sin(x/2))^4`

The form of the answer now suggests a `u`-substitution of `u=cos(x/2)+sin(x/2)`, `du=(\frac{1}{2}cos(x/2)-\frac{1}{2}sin(x/2))\ dx`

However, it seems unclear to me how to take the `sin(x)` in the original numerator and rewrite it in such a way as to get to the final answer, at least without knowing the final answer ahead of time (though I suppose I've already crossed that bridge by using Wolfram Alpha in the first place).

I'll keep working on it and would encourage others to keep working on it.

Answer 2

I'm trying to work backwards by differentiating the answer with the Quotient Rule to get some insight into how to rewrite the `sin(x)` in the numerator of the integrand, but the algebra that arises from using various power-reduction formulas (http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Power-reduction_formula) seems too nasty for me to bother with anymore without technology.

If someone else has a better idea or is willing to go through the algebra in some slick way, it would be great to see.