Question

$\int \frac{1}{7+5\cos x}dx=$

(a) $\frac{1}{\sqrt{6}}{\tan }^{-1}\left(\frac{1}{\sqrt{6}}\tan \frac{x}{2}\right)+C$

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

(c) $\frac{1}{4}{\tan }^{-1}\left(\tan \frac{x}{2}\right)+C$

(d) $\frac{1}{7}{\tan }^{-1}\left(\tan \frac{x}{2}\right)+C$

Answer 1

(a) $\frac{1}{\sqrt{6}}{\tan }^{-1}\left(\frac{1}{\sqrt{6}}\tan \frac{x}{2}\right)+C$

$$\begin{gathered} \mathrm{Let}I=\int \frac{dx}{7+5\cos x} \\ \mathrm{Putting}\cos x=\frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}} \\ \therefore I=\int \frac{dx}{7+5\times \left({\displaystyle \frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}\right)} \\ =\int \frac{\left(1+{\tan }^2{\displaystyle \frac{x}{2}}\right)dx}{7\left(1+{\tan }^2{\displaystyle \frac{x}{2}}\right)+5-5{\tan }^2{\displaystyle \frac{x}{2}}} \\ =\int \frac{{\sec }^2{\displaystyle \frac{\mathrm{x}}{2}}\mathrm{dx}}{2{\tan }^2{\displaystyle \frac{x}{2}}+12} \\ =\frac{1}{2}\int \frac{{\sec }^2{\displaystyle \frac{x}{2}}dx}{{\tan }^2{\displaystyle \frac{x}{2}}+{\left(\sqrt{6}\right)}^2} \\ \mathrm{Let}\tan \frac{x}{2}=t \\ \Rightarrow \frac{1}{2}{\sec }^2\left(\frac{\mathrm{x}}{2}\right)dx=dt \\ \Rightarrow {\sec }^2\left(\frac{\mathrm{x}}{2}\right)dx=2\mathit{}dt \\ \therefore I=\frac{1}{2}\int \frac{2dt}{t^2+{\left(\sqrt{6}\right)}^2} \\ =\frac{1}{\sqrt{6}}{\tan }^{-1}\left(\frac{t}{\sqrt{6}}\right)+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\because \int \frac{1}{a^{\mathit{2}}+x^{\mathit{2}}}\mathit{=}\frac{1}{a}{\tan }^{-1}\frac{x}{a}+C\right) \\ =\frac{1}{\sqrt{6}}{\tan }^{-1}\left(\frac{\tan {\displaystyle \frac{x}{2}}}{\sqrt{6}}\right)+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathit{\because }\mathit{}t=\tan \frac{x}{\mathit{2}}\right) \end{gathered}$$