Evaluate : 1 2-3 2 x d x - Vidyadip

Question

Evaluate : $\int \frac{1}{2-3 \sin 2 x} \cdot d x$

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Answer

put $\tan x=t$
$
\begin{aligned}
& \therefore \quad d x=\frac{1}{1+t^2} \cdot d t \text { and } \sin 2 x=\frac{2}{1+t^2} \\
& =\int \frac{1\left(\frac{1}{1+t^2}\right)}{2-3\left(\frac{2}{1+t^2}\right)} \cdot d t \\
& =\int \frac{\frac{1}{1+t^2}}{\frac{2\left(1+t^2\right)-3(2 t)}{1+t^2}} \cdot d t \\
& =\int \frac{1}{2+2 t^2-6 t} \cdot d t=\int \frac{1}{2\left(t^2-3 t+1\right)} \cdot d t \\
& \because\left\{\left(\frac{1}{2} \text { coefficient of } t\right)^2\right. \\
& \left.=\left(\frac{1}{2}(-3)\right)^2=\left(-\frac{3}{2}\right)^2=\frac{9}{4}\right\} \\
& =\frac{1}{2} \cdot \int \frac{1}{t^2-3 t+\frac{9}{4}-\frac{9}{4}+1} \cdot d t \\
& =\frac{1}{2} \cdot \int \frac{1}{\left(t^2-3 t+\frac{9}{4}\right)-\frac{5}{4}} \cdot d t \\
& =\frac{1}{2} \cdot \int \frac{1}{\left(t-\frac{3}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2} \cdot d t \\
& =\frac{1}{2} \cdot \frac{1}{2\left(\frac{\sqrt{5}}{2}\right)} \cdot \log \left(\frac{\left(t-\frac{3}{2}\right)-\frac{\sqrt{5}}{2}}{\left(t-\frac{3}{2}\right)+\frac{\sqrt{5}}{2}}\right)+c \\
& =\frac{1}{2 \sqrt{5}} \cdot \log \left(\frac{2 t-3-\sqrt{5}}{2 t-3+\sqrt{5}}\right)+c \\
& =\frac{1}{2 \sqrt{5}} \cdot \log \left(\frac{2 \tan x-3-\sqrt{5}}{2 \tan x-3+\sqrt{5}}\right)+c \\
&
\end{aligned}
$

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