Question

Solving integrals of the form $\int \frac{\mathrm{dx}}{\sqrt{x^2+a^2}}$ requires substitution of x =

• A. $a\sec \left(\theta \right)$
• B. $a\tan \left(\theta \right)$
• C. $x=a\mathrm{cosec}\left(\theta \right)$

Answer 1

The correct option is B $a\tan \left(\theta \right)$

Here, we assume $x=a\tan \left(\theta \right)$, then we get,$\mathrm{dx}=a{\sec }^2\theta \mathrm{d\theta }$, now substituting this $x\mathrm{and}\mathrm{dx}$ in the integral we get,
Integral $I=\int \frac{a{\sec }^2\theta \mathrm{d\theta }}{\sqrt{a^2+a^2{\tan }^2\theta }}$
$\Rightarrow I=\int \mathrm{sec\theta }=\ln (\mathrm{tan\theta }+\mathrm{sec\theta })+C$
Now substituting back for $\mathrm{tan\theta }=\frac{x}{a}\mathrm{and}\mathrm{sec\theta }=\frac{\sqrt{x^2+a^2}}{a}$
We get $I=\ln \left(x+\sqrt{x^2+a^2}\right|+C$.