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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
Question
Find $\int \sqrt{3-2 x-x^{2}} d x$
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Answer

Let I = $\int \sqrt{3-2 x-x^{2}} \;d x$
= $\int \sqrt{-\left(x^{2}+2 x-3\right)} d x$
$=\int \sqrt{-\left(x^{2}+1+2 x-3-1\right)} d x$
= $\int \sqrt{-\left[(x+1)^{2}-4\right]} d x$
= $\int \sqrt{4-(x+1)^{2}} d x$
= $\int \sqrt{2^{2}-(x+1)^{2}} d x$ $\left[\because\int\sqrt{a^2-x^2}dx=\frac x2\sqrt{a^2-x^2}+\frac{a^2}2\sin^{-1}\left(\frac xa\right)+c\right]$
= $\frac{1}{2}(x+1) \sqrt{4-(x+1)^{2}} +\frac{4}{2} \sin ^{-1}\left(\frac{x+1}{2}\right)+c$
$\therefore \ \mathrm{I}=\int \sqrt{2^{2}-(x+1)^{2}} d x$
$=\frac{1}{2}(x+1) \sqrt{3-2 x-x^{2}}+2 \sin ^{-1} \frac{(x+1)}{2}+c$

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