Find $$\int \frac{\sin(x)}{1+\sin(x)\cos(x)} dx$$

What I have tried

First method was to try $u$ substitution

Let $u=\cos(x)$ then $-du=\sin(x)dx$ then $\sin(x)=\sqrt{1-u^2} $ which transforms our integral into

$$ \int \frac{\sin(x)}{1+\sin(x)\cos(x)} dx=-\int \frac{1}{1+u\sqrt{1-u^2}}du=-\int \frac{1-u\sqrt{1-u^2}}{u^4-u^2+1}du$$

I think the might the denominator could be seperated since we get $u^2= \frac{1\pm \sqrt{3}i}{2}$ but I wouldn't know how to proceed after that.

Another method I have tried is this

Using the Weierstrass substitution

Let $\tan(x/2)=t$ , so $\sin(t) = \frac{2t}{1+t^2}$ , $\cos(t)=\frac{1-t^2}{1+t^2}$ and $dx=\frac{2dt}{1+t^2}$

Which transforms our integral as such

$$ \int \frac{\sin(x)}{1+\sin(x)\cos(x)}dx= \frac{4t}{t^4-2t^3+2t^2+2t+1} dt$$

I can't seem to find any roots by rational root theorem for the denonminator so I don't know how to proceed once again..

  • $\begingroup$ You have to find the indefinite integral, or are there bounds? $\endgroup$ – GFauxPas Dec 8 '16 at 0:54
  • $\begingroup$ @GFauxPas indefinite. $\endgroup$ – bigfocalchord Dec 8 '16 at 0:55

I wanted to post this as a comment, but I don't have enough points.

Anyway, I guess the trigonometric identity $\sin 2x = 2 \sin x \cdot \cos x$ may be useful for evaluating this integral. Good luck.


\begin{align*} &~~~~~\int\frac{\sin x}{1+\sin x\cos x}{\rm d}x\\ &=\int\frac{2\sin x}{2+2\sin x\cos x}{\rm d}x\\ &=\int\frac{(\sin x+\cos x)+(\sin x-\cos x)}{2+2\sin x\cos x}{\rm d}x\\ &=\int\frac{\sin x+\cos x}{2+2\sin x\cos x}{\rm d}x+\int\frac{\sin x-\cos x}{2+2\sin x\cos x}{\rm d}x\\ &=\int\frac{\sin x+\cos x}{3-(\sin x-\cos x)^2}{\rm d}x+\int\frac{\sin x-\cos x}{(\sin x+\cos x)^2+1}{\rm d}x\\ &=\int\frac{{\rm d}(\sin x-\cos x)}{3-(\sin x-\cos x)^2}-\int\frac{{\rm d}(\sin x+\cos x)}{(\sin x+\cos x)^2+1}\\ &=\frac{1}{2\sqrt{3}}\ln\left|\frac{\sin x-\cos x+\sqrt{3}}{\sin x-\cos x-\sqrt{3}}\right|-\arctan(\sin x+\cos x)+C \end{align*}


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