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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int\frac{\text{x}^2}{(\text{a}-\text{x}^2)^{\frac{3}{2}}}\text{ dx}$
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Answer

Let $\text{I}=\int\frac{\text{x}^2}{(\text{a}-\text{x}^2)^{\frac{3}{2}}}\text{ dx}$
Let $\text{x}=\text{a}\cos\theta$
On differentiating both sides, we get
$\text{dx}=-\text{a}\sin\theta\text{ d}\theta$
$\therefore\ \text{I}=\int\frac{\text{a}^2\cos^2\theta}{(\text{a}^2-\text{a}^2\cos^2\theta)^\frac{3}{2}}\times-\text{a}\sin\theta\text{ d}\theta$
$=-\int\frac{\text{a}^3\cos^2\theta\sin\theta}{\text{a}^3(1-\cos^2\theta)^{\frac{3}{2}}}\text{ d}\theta$
$=-\int\frac{\cos^2\theta\sin\theta}{\sin^3\theta}\text{ d}\theta$
$=-\int\cot^2\theta\text{ d}\theta$
$=-\int(\text{cosec}^2\theta-1)\text{d}\theta$
$=-(-\cot\theta-\theta)+\text{C}$
$=\cot\theta+\theta+\text{C}$
$=\cot\Big(\cos^{-1}\frac{\text{x}}{\text{a}}\Big)+\cos^{-1}\frac{\text{x}}{\text{a}}+\text{C}$
$=\cot\Big(\cos^{-1}\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\Big)+\cos^{-1}\frac{\text{x}}{\text{a}}+\text{C}$
$=\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}+\cos^{-1}\frac{\text{x}}{\text{a}}+\text{C}$
Hence, $\int\frac{\text{x}^2}{(\text{a}^2-\text{x}^2)^{\frac{3}{2}}}\text{ dx}=\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}+\cos^{-1}\frac{\text{x}}{\text{a}}+\text{C}$

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