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Indefinite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integration4 Marks
Question
Evaluate : $\int \sqrt{4+3 x-2 x^2} \cdot d x$
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Answer

$
\begin{gathered}
\mathrm{I}=\int \sqrt{4-2 x^2+3 x} \cdot d x \\
=\int \sqrt{4-2\left(x^2-\frac{3}{2} x\right)} \cdot d x \\
=\int \sqrt{2} \cdot \sqrt{2-\left(x^2-\frac{3}{2} x\right)} \cdot d x
\end{gathered}
$
$
\because \quad\left\{\left(\frac{1}{2} \text { coefficient of } x\right)^2=\left[\frac{1}{2}\left(-\frac{3}{2}\right)\right]^2=\left(-\frac{3}{4}\right)^2=\frac{9}{16}\right\}
$
$
\begin{aligned}
& \mathrm{I}=\sqrt{2} \cdot \int \sqrt{2-\left(x^2-\frac{3}{2} x+\frac{9}{16}-\frac{9}{16}\right)} \cdot d x \\
& =\sqrt{2} \cdot \int \sqrt{2-\left(x^2-\frac{3}{2} x+\frac{9}{16}\right)+\frac{9}{16}} \cdot d x \\
& =\sqrt{2} \cdot \int \sqrt{\left(\frac{\sqrt{41}}{4}\right)^2-\left(x-\frac{3}{4}\right)^2} \cdot d x \\
& \because \int \sqrt{a^2-x^2} \cdot d x=\frac{x}{2} \cdot \sqrt{a^2-x^2}+\frac{a^2}{2} \cdot \sin ^{-1}\left(\frac{x}{a}\right)+c \\
& =\sqrt{2} \cdot\left[\frac{\left(x-\frac{3}{4}\right)}{2} \cdot \sqrt{\left(\frac{\sqrt{41}}{4}\right)^2-\left(x-\frac{3}{4}\right)^2}+\frac{\left(\frac{\sqrt{41}}{4}\right)^2}{2} \cdot \sin ^{-1}\left(\frac{x-\frac{3}{4}}{\frac{\sqrt{41}}{4}}\right)\right]+c \\
& =\sqrt{2}\left[\frac{4 x-3}{8} \cdot \sqrt{2+\frac{3}{2} x-x^2}+\frac{41}{32} \cdot \sin ^{-1}\left(\frac{4 x-3}{\sqrt{41}}\right)\right]+c \\
& \therefore \int \sqrt{4+3 x-2 x^2} \cdot d x=\frac{4 x-3}{8} \cdot \sqrt{4+3 x-2 x^2}+\frac{41}{16 \sqrt{2}} \cdot \sin ^{-1}\left(\frac{4 x-3}{\sqrt{41}}\right)+c \\
&
\end{aligned}
$

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