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(C) Let $I = \int \frac{dx}{(x^2 - a^2)^{3/2}}$.Substitute $x = a \sec 	heta$, then $dx = a \sec 	heta 	an 	heta d	heta$.Substituting these into

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$-\frac{x}{a^2 \sqrt{x^2-a^2}}+C$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{(x^2 - a^2)^{3/2}}$.Substitute $x = a \sec 	heta$, then $dx = a \sec 	heta 	an 	heta d	heta$.Substituting these into the integral:$I = \int \frac{a \sec 	heta 	an 	heta d	heta}{(a^2 	an^2 	heta)^{3/2}} = \int \frac{a \sec 	heta 	an 	heta d	heta}{a^3 	an^3 	heta} = \frac{1}{a^2} \int \frac{\sec 	heta}{	an^2 	heta} d	heta$.Using $\sec 	heta = \frac{1}{\cos 	heta}$ and $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$, we get:$I = \frac{1}{a^2} \int \frac{1}{\cos 	heta} \cdot \frac{\cos^2 	heta}{\sin^2 	heta} d	heta = \frac{1}{a^2} \int \frac{\cos 	heta}{\sin^2 	heta} d	heta$.Let $u = \sin 	heta$, then $du = \cos 	heta d	heta$. So,$I = \frac{1}{a^2} \int u^{-2} du = \frac{1}{a^2} (-u^{-1}) + C = -\frac{1}{a^2 \sin 	heta} + C$.Since $x = a \sec 	heta$, $\sec 	heta = \frac{x}{a}$, so $\cos 	heta = \frac{a}{x}$.Then $\sin 	heta = \sqrt{1 - \cos^2 	heta} = \sqrt{1 - \frac{a^2}{x^2}} = \frac{\sqrt{x^2 - a^2}}{x}$.Substituting $\sin 	heta$ back, we get:$I = -\frac{1}{a^2 \left( \frac{\sqrt{x^2 - a^2}}{x} ight)} + C = -\frac{x}{a^2 \sqrt{x^2 - a^2}} + C$.

$\int \frac{d x}{\left(x^2-a^2\right)^{\frac{3}{2}}}$ is equal to

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