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

Question

Scarlett Allison · Accepted Answer

This is one of the most difficult integrals I've seen in a long time. Given that I know the solution from Wolfram Alpha, I believe I have a good start on it. Can someone assist me in finishing it?

Wolfram Alpha generates the following: https://tutor.hix.ai

#\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 continue to work on it, and I'd advise others to do the same.

Charles Anctil · Answer

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 (https://tutor.hix.ai) seems too nasty for me to bother with anymore without technology. It would be fantastic to see if someone else could solve the algebra more cleverly or has a better idea.

Cameron Alexander · Answer

undefined To integrate $\frac{\sin x}{(1+\sin x)^2}$, you can use the substitution method. Let $u = 1 + \sin x$, then $du = \cos x dx$. After substitution, the integral becomes $\int \frac{1}{u^2} du$, which is straightforward to integrate. The final result will be $-\frac{1}{1+\sin x} + C$, where $C$ is the constant of integration.