# How do you test the improper integral int sintheta/sqrtcostheta from [0,pi/2] and evaluate if possible?

Apr 13, 2017

$2$

#### Explanation:

$I = {\int}_{0}^{\frac{\pi}{2}} \sin \frac{\theta}{\sqrt{\cos}} \theta d \theta$

Use the substitution $u = \cos \theta$. This implies that $\mathrm{du} = - \sin \theta d \theta$.

When we substitute this into the integral, we will have to transform the bounds by plugging the current ones into $\cos \theta$, so the bound of $0$ will become $\cos \left(0\right) = 1$ and the bound of $\frac{\pi}{2}$ will become $\cos \left(\frac{\pi}{2}\right) = 0$.

Then:

$I = - {\int}_{0}^{\frac{\pi}{2}} \frac{- \sin \theta}{\sqrt{\cos}} \theta d \theta = - {\int}_{1}^{0} \frac{1}{\sqrt{u}} \mathrm{du}$

Flipping the integral's bounds with the negative sign and rewriting the exponent:

$I = {\int}_{0}^{1} {u}^{- \frac{1}{2}} \mathrm{du}$

Using the rule $\int {u}^{n} \mathrm{du} = {u}^{n + 1} / \left(n + 1\right) + C$ but using the FTC to evaluate the integral:

$I = {\left[{u}^{\frac{1}{2}} / \left(\frac{1}{2}\right)\right]}_{0}^{1} = {\left[2 \sqrt{u}\right]}_{0}^{1} = 2 \sqrt{1} - 2 \sqrt{0} = 2$