Evaluate : 1 5-4 x d x - Vidyadip

Question

Evaluate : $\int \frac{1}{5-4 \cos x} \cdot d x$

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Answer

put $\tan \frac{x}{2}=t$
$
\begin{aligned}
& \therefore \quad d x=\frac{2}{1+t^2} \cdot d t \text { and } \quad \cos x=\frac{1-t^2}{1+t^2} \\
& I=\int \frac{1\left(\frac{2}{1+t^2}\right)}{5-4\left(\frac{1-t^2}{1+t^2}\right)} \cdot d t \\
& =\int \frac{\frac{2}{1+t^2}}{\frac{5\left(1+t^2\right)-4\left(1-t^2\right)}{1+t^2}} \cdot d t \\
& =\int \frac{2}{5-5 t^2-4-4 t^2} \cdot d t \\
& =\int \frac{2}{9 t^2+1} \cdot d t \\
& =\int \frac{1}{9\left(t^2+\frac{1}{9}\right)} \cdot d t \\
& =\frac{2}{9} \cdot \int \frac{1}{t^2+\left(\frac{1}{3}\right)^2} \cdot d t \\
& =\frac{2}{9} \cdot \frac{1}{\left(\frac{1}{3}\right)} \cdot \tan ^{-1}\left(\frac{t}{\left(\frac{1}{3}\right)}\right)+c \\
& =\frac{2}{3} \cdot \tan ^{-1}(2 t)+c \\
& =\frac{2}{3} \cdot \tan ^{-1}\left(2 \tan \frac{x}{2}\right)+c \\
& \therefore \int \frac{1}{5-4 \cos x} \cdot d x=\frac{2}{3} \cdot \tan ^{-1}\left(2 \tan \frac{x}{2}\right)+c \\
\end{aligned}
$

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