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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS3 Marks
Question
Integrate the function in Exercise:
$\frac{1}{\text{x}\sqrt{\text{ax}-\text{x}^{2}}}$
$\big[\text{Hint:putx}=\frac{\text{a}}{\text{t}}\big]$
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Answer

$\frac{1}{\text{x}\sqrt{\text{ax}-\text{x}^{2}}}$
$\text{Let}\ \text{x}=\frac{\text{a}}{\text{t}}\Rightarrow\text{dx}=-\frac{\text{a}}{\text{t}^{2}}\text{dt}$
$\Rightarrow\int\frac{1}{\text{x}\sqrt{\text{ax}-\text{x}^{2}}}\text{dx}=\int\frac{1}{\frac{\text{a}}{\text{t}}\sqrt{\text{a}.\frac{\text{a}}{\text{t}}-\big(\frac{\text{a}}{\text{t}}}\big)^{2}}\Big(-\frac{\text{a}}{\text{t}^{2}}\text{dt}\Big)$
$=-\int\frac{1}{\text{at}}.\frac{1}{\sqrt{\frac{1}{\text{t}}-\frac{1}{\text{t}^{2}}}}\text{dt}$
$=-\frac{1}{\text{a}}\int\frac{1}{\sqrt{\frac{\text{t}^{2}}{\text{t}}-\frac{\text{t}^{2}}{\text{t}^{2}}}}\text{dt}$
$=-\frac{1}{\text{a}}\int\frac{1}{\sqrt{\text{t}-1}}\text{dt}$
$=-\frac{1}{\text{a}}\big[2\sqrt{\text{t}-1}\big]+\text{C}$
$=-\frac{1}{\text{a}}\bigg[2\sqrt{\frac{\text{a}}{\text{x}}-1}\bigg]+\text{C}$
$=-\frac{2}{\text{a}}\bigg(\frac{\sqrt{\text{a}-\text{x}}}{\sqrt{\text{x}}}\bigg)+\text{C}$
$=-\frac{2}{\text{a}}\bigg(\sqrt{\frac{\text{a}-\text{x}}{\text{x}}}\bigg)+\text{C}$

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