Evaluate the following integrals: x^7 (a^2-x^2)^5 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{\text{x}^7}{(\text{a}^2-\text{x}^2)^5}\text{ dx}$

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Answer

Let $\text{I}=\int\frac{\text{x}^7}{(\text{a}^2-\text{x}^2)^5}\text{ dx}$
Let $\text{x}=\text{a}\sin\theta$
On differentiating both sides, we get
$\text{dx}=\text{x}\cos\theta\text{ d}\theta$
$\therefore\ \text{I}=\int\frac{\text{a}^8\sin^{7}\theta\cos\theta}{(\text{a}^2-\text{a}^2\sin^{2}\theta)^5}\text{ d}\theta$
$=\int\frac{\text{a}^8\sin^{7}\theta\cos\theta}{\text{a}^{10}(1-\sin^2\theta)^5}\text{ d}\theta$
$=\int\frac{\sin^7\theta}{\text{a}^2\cos^9\theta}\text{ d}\theta$
$=\frac{1}{\text{a}^2}\int\tan^7\theta\sec^2\theta\text{ d}\theta$
Let $\tan\theta=\text{t}$
On differentiating both sides, we get
$\sec^2\theta\text{ d}\theta=\text{dt}$
$\therefore\ \text{I}=\frac{1}{\text{a}^2}\int\text{t}^7\text{dt}$
$=\frac{1}{\text{a}^2}\frac{\text{t}^8}{8}+\text{C}$
$=\frac{1}{8\text{a}^2}(\tan^8\theta)+\text{C}$
$=\frac{1}{8\text{a}^2}\Big(\tan\Big(\sin^{-1}\frac{\text{x}}{\text{a}}\Big)\Big)^8+\text{C}$
$=\frac{1}{8\text{a}^2}\Big(\tan\Big(\tan^{-1}\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\Big)\Big)^8+\text{C}$
$=\frac{1}{8\text{a}^2}\frac{\text{x}^8}{(\text{a}^2-\text{x}^2)^4}+\text{C}$
Hence, $\int\frac{\text{x}^7}{(\text{a}^2-\text{x}^2)^5}\text{ dx}=\frac{1}{8\text{a}^2}\frac{\text{x}^8}{(\text{a}^2-\text{x}^2)^4}+\text{C}$

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