Question

$\int \frac{\sin x}{3+4{\cos }^{2}x}dx=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

$I=\int \frac{\sin x}{3+4{\cos }^{2}x}dx$

Let $\cos x=z$

$\Rightarrow -\sin xdx=dz$

$\therefore I=-\int \frac{dz}{3+4z^2}$

$\Rightarrow I=-\frac{1}{4}\int \frac{dz}{z^2+{\displaystyle \frac{3}{4}}}$

$\Rightarrow I=-\frac{1}{4}\int \frac{dz}{z^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2}$

$\Rightarrow I=-\frac{1}{4}\times \frac{1}{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}{\tan }^{-1}\left(\frac{z}{{\displaystyle \frac{\sqrt{3}}{2}}}\right)+C\left[\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C\right]$

$\Rightarrow I=-\frac{1}{2\sqrt{3}}{\tan }^{-1}\left(\frac{2\cos x}{\sqrt{3}}\right)+C$


$\int \frac{\sin x}{3+4{\cos }^{2}x}dx=$$\underline{-\frac{1}{2\sqrt{3}}{\tan }^{-1}\left(\frac{2\cos x}{\sqrt{3}}\right)+C}$