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Indefinite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integration5 Marks
Question
Evaluate : $\int \frac{\sin 2 x}{3 \sin ^4 x-4 \sin ^2 x+1} \cdot d x$
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Answer

$\mathrm{I}=\int \frac{\sin 2 x}{3\left(\sin ^2 x\right)^2-4\left(\sin ^2 x\right)+1} \cdot d x$ put $\sin ^2 x=t$ $\therefore 2 \sin x \cdot \cos x \cdot d x=1 \cdot d t$ $\therefore \quad \sin 2 x \cdot d x=1 \cdot d t$
$=\int \frac{1}{3 t^2-4 t+1} \cdot d t$
$=\int \frac{1}{3\left(t^2-\frac{4}{3} t+\frac{1}{3}\right)} \cdot d t$
$\because\left\{\left(\frac{1}{2} \text { coefficient of } t\right)^2\right.$
$\left.=\left(\frac{1}{2}\left(-\frac{4}{3}\right)\right)^2=\left(-\frac{2}{3}\right)^2=\frac{4}{9}\right\}$
$I=\frac{1}{3} \cdot \int \frac{1}{t^2-\frac{4}{3} t+\frac{4}{9}-\frac{4}{9}+\frac{1}{3}} \cdot d t$
$=\frac{1}{3} \cdot \int \frac{1}{\left(t^2-\frac{4}{3} t+\frac{4}{9}\right)-\frac{1}{9}} \cdot d t$
$=\frac{1}{3} \cdot \int \frac{1}{\left(t-\frac{2}{3}\right)^2-\left(\frac{1}{3}\right)^2} \cdot d t$
$=\frac{1}{3} \cdot \frac{1}{2\left(\frac{1}{3}\right)} \cdot \log \left(\frac{\left(t-\frac{2}{3}\right)-\frac{1}{3}}{\left(t-\frac{2}{3}\right)+\frac{1}{3}}\right)+c$
$=\frac{1}{2} \cdot \log \left(\frac{3 t-3}{3 t-1}\right)+c$
$=\frac{1}{2} \cdot \log \left(\frac{3 \sin ^2 x-3}{3 \sin ^2 x-1}\right)+c$
$\therefore \quad \int \frac{\sin 2 x}{3 \sin ^4 x-4 \sin ^2 x+1} \cdot d x$
$=\frac{1}{2} \cdot \log \left(\frac{3 \sin ^2 x-3}{3 \sin ^2 x-1}\right)+c$

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