# How do you integrate int 1/sqrt(25-t^2) by trigonometric substitution?

Mar 19, 2017

$\int \frac{\mathrm{dt}}{\sqrt{25 - {t}^{2}}} = \arcsin \left(\frac{t}{5}\right) + C$

#### Explanation:

Substitute:

$t = 5 \sin x$
$\mathrm{dt} = 5 \cos x$

so that:

$\int \frac{\mathrm{dt}}{\sqrt{25 - {t}^{2}}} = \int \frac{5 \cos x \mathrm{dx}}{\sqrt{25 - 25 {\sin}^{2} x}}$

$\int \frac{\mathrm{dt}}{\sqrt{25 - {t}^{2}}} = \int \frac{5 \cos x \mathrm{dx}}{\sqrt{25 \left(1 - {\sin}^{2} x\right)}}$

$\int \frac{\mathrm{dt}}{\sqrt{25 - {t}^{2}}} = \int \frac{\cancel{5} \cos x \mathrm{dx}}{\cancel{5} \sqrt{{\cos}^{2} x}}$

Now, the integrand function is defined only for $t \in \left(- 5 , 5\right)$, so that we can assume $x \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$. In this interval $\cos x$ is always positive, so:

$\sqrt{{\cos}^{2} x} = \cos x$

$\int \frac{\mathrm{dt}}{\sqrt{25 - {t}^{2}}} = \int \frac{\cos x \mathrm{dx}}{\cos x} = \int \mathrm{dx} = x + C$

Undoing the substitution:

$\int \frac{\mathrm{dt}}{\sqrt{25 - {t}^{2}}} = \arcsin \left(\frac{t}{5}\right) + C$