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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
Question
Integrate the function $\frac{1}{x \sqrt{a x-x^{2}}}$ [Hint: Put x = $\frac{a}{t}$]
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Answer

Given: $\frac{1}{x \sqrt{a x-x^{2}}}$
Let $I=\frac{1}{x \sqrt{a x-x^{2}}}$
Put $x=\frac{a}{t} \Rightarrow d x=-\frac{a}{t^{2}} d t$
$\Rightarrow \int \frac{1}{x \sqrt{a x-x^{2}}} d x=\int \frac{1}{\frac{a}{t} \sqrt{\frac{a \cdot a}{t}-\left(\frac{a}{t}\right)^{2}}} \cdot-\frac{a}{t^{2}} d t$
$=\int \frac{-1}{a t} \cdot \frac{1}{\sqrt{\frac{1}{t}-\left(\frac{1}{t}\right)^{2}}} d t$
= $-\frac{1}{a} \int \frac{1}{\sqrt{\frac{t^{2}}{t}-\left(\frac{t}{t}\right)^{2}}} d t$
= $-\frac{1}{a} \int \frac{1}{\sqrt{t-1}} d t$
= $-\frac{1}{a} \int(t-1)^{-\frac{1}{2}} d t$
= $-\frac{1}{a}\left[\frac{\sqrt{(t-1)}}{\frac{1}{2}}\right]+C$
= $-\frac{2}{a}[\sqrt{\left(\frac{a}{x}-1\right)}]+C$ ; because $t=\frac{a}{x}$
$\Rightarrow \mathrm{I}=-\frac{2}{\mathrm{a}}[\sqrt{\left(\frac{\mathrm{a}-\mathrm{x}}{\mathrm{x}}\right)}]+\mathrm{C}$

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