Question

$\int \frac{1}{\sqrt{a^2 - x^2}} \, dx =$

• A. $\log |x + \sqrt{x^2 + a^2}| + c$
• B. $\log |x + \sqrt{x^2 - a^2}| + c$
• C. $\sin^{-1} \frac{x}{a} + c$
• D. $\frac{1}{a} \sin^{-1} \frac{x}{a} + c$

Hint:

Don't confuse this with $\int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1} \frac{x}{a} + c$. Only the $\tan^{-1}$ and $\sec^{-1}$ forms have a $1/a$ coefficient in front; the $\sin^{-1}$ form does not.

Answer 1

The Correct Option is C

Step 1: Choose Substitution: Let $x = a \sin \theta$. Then $dx = a \cos \theta \, d\theta$. Also, $\sin \theta = \frac{x}{a} \implies \theta = \sin^{-1} \frac{x}{a}$.

Step 2: Substitute into the Integral: $$\int \frac{a \cos \theta}{\sqrt{a^2 - (a \sin \theta)^2}} \, d\theta$$ $$\int \frac{a \cos \theta}{\sqrt{a^2(1 - \sin^2 \theta)}} \, d\theta$$

Step 3: Simplify: Using the identity $1 - \sin^2 \theta = \cos^2 \theta$: $$\int \frac{a \cos \theta}{\sqrt{a^2 \cos^2 \theta}} \, d\theta$$ $$\int \frac{a \cos \theta}{a \cos \theta} \, d\theta = \int 1 \, d\theta\lt strong\gt Step 4: Integrate and Back-Substitute\lt /strong\gt \int 1 \, d\theta = \theta + c$$ Substitute $\theta = \sin^{-1} \frac{x}{a}$: $$\text{Integral} = \sin^{-1} \frac{x}{a} + c$$