$\int \frac{d x}{7+5 \cos x}$ is equal to
$\int \frac{d x}{7+5 \cos x}$ is equal to
- $\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{1}{\sqrt{3}} \tan \frac{x}{2}\right)+c$
- $\frac{1}{\sqrt{6}} \tan ^{-1}\left(\frac{1}{\sqrt{6}} \tan \frac{x}{2}\right)+c$
- $\frac{1}{7} \tan ^{-1}\left(\tan \frac{x}{2}\right)+c$
- $\frac{1}{4} \tan ^{-1}\left(\tan \frac{x}{2}\right)+c$
Solution
$
\text { Let } \begin{aligned}
I & =\int \frac{d x}{7+5 \cos x} \\
& =\int \frac{d x}{7\left(\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}\right)+5\left(\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}\right)} \\
& =\int \frac{d x}{12 \cos ^2 \frac{x}{2}+2 \sin ^2 \frac{x}{2}} \\
& =\frac{1}{2} \int \frac{\sec ^2 \frac{x}{2}}{6+\tan ^2 \frac{x}{2}} d x
\end{aligned}
$
Let $\tan \frac{x}{2}=z$
$
\begin{aligned}
& \Rightarrow \frac{1}{2} \sec ^2 \frac{x}{2} d x=d z \\
& \therefore \quad I=\int \frac{d z}{6+z^2} \\
& =\frac{1}{\sqrt{6}} \tan ^{-1} \frac{z}{\sqrt{6}}+c \\
& =\frac{1}{\sqrt{6}} \tan ^{-1}\left(\frac{1}{\sqrt{6}} \tan \frac{x}{2}\right)+c \\
&
\end{aligned}
$
Asked in: AP EAMCET 2002
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