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

Question

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..

Answer 1

\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*}

Answer 2

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.

Answer 3

Notice that

$$\left(t^4 - 2t^3 + 2t^2 + 2t + 1\right) \left(t^4 \color{red}+ 2t^3 + 2t^2 \color{red}- 2t + 1\right) = t^8 + 14t^4 + 1$$

which has roots (assuming the principal branch of $\sqrt z$) at

$$t^4=-7\pm4\sqrt3 \implies t^2 = i \left(2\pm\sqrt3\right) \implies t = \sqrt{2\pm\sqrt3} \, e^{i\theta}, \theta \in \left\{\pm\dfrac{3\pi}4, \pm\dfrac\pi4\right\}$$

From this we can infer the roots and subsequent factorization of

$$\begin{align*} f(t) &= t^4 - 2t^3 + 2t^2 + 2t + 1 \\ &= \left(t^2 + (1-a) t + a\right) \left(t^2 + (1-b) t + b\right) \end{align*}$$

where $(a,b) = \left(2-\sqrt3, 2+\sqrt3\right)$.

Partial fraction expansion is the next logical step; we find

$$\frac{4t}{f(t)} = \frac{At+B}{t^2+(1-a)t+a} + \frac{Ct+D}{t^2+(1-b)t+b} \implies (A,B,C,D) = \frac{(1,-a,-1,b)}{\sqrt3}$$

which leaves us with the easy but unexciting task of recovering the same (up to constant of integration) $\log$ and $\arctan$ expressions given in mengdie1982's answer.