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

Aug 14, 2017

Use limits.

#### Explanation:

The integral is improper where $\cos \theta = 0$, which occurs at $\theta = \frac{\pi}{2}$.
To evaluate,
${\int}_{0}^{\frac{\pi}{2}} \sin \frac{\theta}{\cos} ^ 2 \theta d \left(\theta\right)$
first observe that $\sin \frac{\theta}{\cos} ^ 2 \theta = \sec \left(\theta\right) \tan \left(\theta\right)$
Use a limit on the improper end:
${\int}_{0}^{\frac{\pi}{2}} \sec \left(\theta\right) \tan \left(\theta\right) d \left(\theta\right) = \lim \left({\int}_{0}^{b} \sec \left(\theta\right) \tan \left(\theta\right) d \left(\theta\right)\right)$, where the limit is as $b \to {\left(\frac{\pi}{2}\right)}^{-}$
Evaluate:
$\lim \left({\int}_{0}^{b} \sec \left(\theta\right) \tan \left(\theta\right) d \left(\theta\right)\right) =$
$= \lim \left(\sec b - \sec 0\right)$
$= \lim \left(\sec b - 1\right)$
This limit diverges as $b \to \frac{\pi}{2}$.