How can I evaluate following integral $$\int\frac{\sin x}{3-2\sin x}dx$$
My first attempt:
Substitute $3-2\sin x=t$ or $\sin x=\frac{3-t}{2}$, $$\cos xdx=-\frac{1}{2}dt, \ dx=-\frac{dt}{\sqrt{6t-t^2-7}}$$$$\int\frac{\sin x}{3-2\sin x}dx=\int\frac{\frac{3-t}{2}}{t}\cdot -\frac{dt}{2\sqrt{6t-t^2-7}}$$$$=-\int\frac{(3-t)dt}{t\sqrt{6t-t^2-7}}$$
Here I got stuck by using this substitution
My second attempt:
I used half angle formula: $\sin x=\frac{2\tan \frac{x}{2}}{1+\tan^{2}\frac{x}{2}}$ which led me
$$\int\frac{\sin x}{3-2\sin x}dx=\int\frac{\frac{2\tan \frac{x}{2}}{1+\tan^{2}\frac{x}{2}}}{3-2\frac{2\tan \frac{x}{2}}{1+\tan^{2}\frac{x}{2}}}$$$$=\int\frac{2\tan \frac{x}{2}}{3+3\tan^2 \frac{x}{2}-4\tan\frac{x}{2}}$$
Denominator made perfect square $$=\frac{2}{3}\int\frac{2\tan \frac{x}{2}}{\left(\tan \frac{x}{2}-\frac23\right)^2-\frac13}$$
I got stuck here, don't know how to proceed.
Please help me solve this integration. Thanks in advance
Best Answer
Use partial fractions then half angle formula as follows $$\int\frac{\sin x}{3-2\sin x}dx=\frac12\int\frac{3-(3-2\sin x)}{3-2\sin x}dx$$
$$=\frac12\int\left(\frac{3}{3-2\sin x}-1\right)dx$$ $$=\frac32\int\frac{dx}{3-2\sin x}-\frac12\int 1dx$$ $$=\frac32\int\frac{dx}{3-2\cdot\frac{2\tan\frac x2}{1+\tan^2\frac x2}}-\frac12x+C$$ $$=\frac12\int\frac{\sec^2\frac x2dx}{\left(\tan\frac x2-\frac23\right)^2-\frac13}-\frac12x+C$$ $$=\int\frac{d\left(\tan\frac x2-\frac23\right)}{\left(\tan\frac x2-\frac23\right)^2-\frac13}-\frac12x+C$$ $$=\frac{1}{2\cdot \frac{1}{\sqrt3}}\ln\left|\frac{\tan\frac x2-\frac23-\frac{1}{\sqrt3}}{\tan\frac x2-\frac23+\frac{1}{\sqrt3}}\right|-\frac12x+C$$ $$=\frac{\sqrt3}{2}\ln\left|\frac{3\tan\frac x2-2-\sqrt3}{3\tan\frac x2-2+\sqrt3}\right|-\frac12x+C$$