Question

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

(a) log (3 + 4 cos² x) + C

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

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

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

Answer 1

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

$$\begin{gathered} \mathrm{Let}I=\int \frac{\sin x}{3+4{\cos }^{2}x}dx \\ \mathrm{Putting}\cos x=t \\ \Rightarrow -\sin xdx=dt \\ \therefore I=\int \frac{-dt}{3+4t^2} \\ =\frac{1}{4}\int \frac{-dt}{t^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2} \\ =\frac{-1}{4}\times \frac{1}{{\displaystyle \frac{\sqrt{3}}{2}}}{\tan }^{-1}\left(\frac{t\times 2}{\sqrt{3}}\right)+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathit{\because }\int \frac{\mathit{1}}{x^{\mathit{2}}+a^{\mathit{2}}}=\frac{1}{a}{\tan }^{\mathit{-}1}\frac{x}{a}+C\right) \\ =-\frac{1}{2\sqrt{3}}{\tan }^{-1}\left(\frac{2t}{\sqrt{3}}\right)+C \\ =-\frac{1}{2\sqrt{3}}{\tan }^{-1}\left(\frac{2\cos x}{\sqrt{3}}\right)+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\because t=\cos \mathit{}x\right) \end{gathered}$$