Find the integral using an appropriate trigonometric substitution.

Question

Quizplus · Accepted Answer

The integral can be solved by using the trigonometric substitution $x=\sqrt{a^2-u^2}$, which gives $dx=-\frac{u}{\sqrt{a^2-u^2}}du$ and $\sqrt{a^2-x^2}=u$. Plugging these into the integral, we get:

$$\int{\frac{dx}{x^2\sqrt{a^2-x^2}}}=\int{\frac{-1/a^2}{(1-(x/a)^2)^{3/2}}dx}$$

Now, let $x/a=\sin	heta$, so $dx=a\cos	heta d	heta$. Substituting this in and simplifying:

$$\int{\frac{-a\cos	heta d	heta}{a^2\sin^2	heta\sqrt{1-\sin^2	heta}}}=-\int{\frac{\cos	heta d	heta}{a\sin^2	heta}}$$

Using the identity $\sin^2	heta=1-\cos^2	heta$, this becomes:

$$-\int{\frac{d(\sin	heta)}{a\sin^2	heta}}=\frac{1}{a\sin	heta}+C=\frac{1}{a\sqrt{1-(x/a)^2}}+C$$

Therefore, the answer is (B).

Find the Integral Using an Appropriate Trigonometric Substitution