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$\int \frac{d x}{4+5 \cos x}=$
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Verified Answer
The correct answer is:
$\frac{1}{3} \log \left|\frac{3+\tan \frac{x}{2}}{3-\tan \frac{x}{2}}\right|+C$
$\begin{aligned}
& \text {Let } I=\int \frac{1}{4+5 \cos x} d x \\
& =\int \frac{1}{4+5\left(\frac{1-\tan ^2 x / 2}{1+\tan ^2 x / 2}\right)} d x
\end{aligned}$
Let $u=\tan \frac{x}{2} \Rightarrow \frac{d u}{d x}=\frac{\sec ^2 \frac{x}{2}}{2} d u$
$\begin{aligned} & =-2 \int \frac{1}{u^2-9} d u=\frac{1}{3} \int\left(\frac{1}{u+3}+\frac{1}{u-3}\right) d u \\ & =\frac{1}{3} \ln \left(\frac{3+u}{3-u}\right)+c=\frac{1}{3} \ln \left(\frac{\tan \frac{x}{2}+3}{3-\tan \frac{x}{2}}\right)+C\end{aligned}$
& \text {Let } I=\int \frac{1}{4+5 \cos x} d x \\
& =\int \frac{1}{4+5\left(\frac{1-\tan ^2 x / 2}{1+\tan ^2 x / 2}\right)} d x
\end{aligned}$
Let $u=\tan \frac{x}{2} \Rightarrow \frac{d u}{d x}=\frac{\sec ^2 \frac{x}{2}}{2} d u$
$\begin{aligned} & =-2 \int \frac{1}{u^2-9} d u=\frac{1}{3} \int\left(\frac{1}{u+3}+\frac{1}{u-3}\right) d u \\ & =\frac{1}{3} \ln \left(\frac{3+u}{3-u}\right)+c=\frac{1}{3} \ln \left(\frac{\tan \frac{x}{2}+3}{3-\tan \frac{x}{2}}\right)+C\end{aligned}$
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