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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int\frac{1}{\sqrt{(\text{x}-\alpha)(\beta-\text{x})}}\text{ dx},(\beta>\alpha)$
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Answer

Let $\text{I}=\int\frac{1}{\sqrt{(\text{x}-\alpha)(\beta-\text{x})}}\text{ dx},(\beta>\alpha)$
$=\int\frac{1}{-\text{x}^2-\text{x}(\alpha+\beta)-\alpha\beta}\text{ dx}$
$=\int\frac{1}{\sqrt{-\Big[\text{x}^2-2\text{x}\big(\frac{\alpha+\beta}{2}\big)+\big(\frac{\alpha+\beta}{2}\big)^2-\big(\frac{\alpha+\beta}{2}\big)^2+\alpha\beta\Big]}}\text{ dx}$
$=\int\frac{1}{\sqrt{-\Big[\big(\text{x}-\frac{\alpha+\beta}{2}\big)^2-\big(\frac{\alpha+\beta}{2}\big)^2\Big]}}\text{ dx}$
$=\int\frac{1}{\sqrt{\Big[\big(\frac{\beta-\alpha}{2}\big)^2-\big(\text{x}-\frac{\alpha+\beta}{2}\big)^2\Big]}}\text{ dx}$ $[\therefore\ \beta>\alpha]$
Let $\Big(\text{x}-\frac{\alpha+\beta}{2}\Big)=\text{t}$
$\Rightarrow\text{dx}=\text{dt}$
$\text{I}=\int\frac{1}{\sqrt{\big(\frac{\beta-\alpha}{2}\big)^2-\text{t}^2}}\text{ dt}$
$\text{I}=\sin^{-1}\bigg(\frac{\text{t}}{\frac{\beta-\alpha}{2}}\bigg)+\text{C}$ $\Big[\text{Since }\int\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\text{ dx}=\sin^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
$\text{I}=\sin^{-1}\Bigg(\frac{2\big(\text{x}-\frac{\alpha+\beta}{2}\big)}{\beta-\alpha}\Bigg)+\text{C}$
$\text{I}=\sin^{-1}\Big(\frac{2\text{x}-\alpha-\beta}{\beta-\alpha}\Big)+\text{C}$

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