Could someone please help me with this? My professor kind of zoomed over explaining this and I'm having problems...
The problem is: Let a>0. Determine if the improper integral $ \int_0^ \frac{ \pi }{2} \sin ^{a}\theta \tan\theta d\theta$ converges.
What I've done so far:
Let $u=\tan\theta, du=\sec ^{2}\theta d\theta$
$ \int_0^ \infty \frac{ \sin ^{a}\theta u}{\sec ^{2}\theta } du $
Now I'm not sure what to do. I tried using (a) $\sin ^{2}\theta = 1-\cos ^{2}\theta \Longrightarrow \sin \theta = \sqrt{1-\cos ^{2}} $
and (b) $u=\tan\theta\Longrightarrow \sec ^{2}\theta=1+\tan ^{2}\theta=1+u^2$
which makes $ \int_0^ \infty \frac{u^2}{(1+u^2)^\frac{3}{2} } du $....?
Am I on the right track?? How do I integrate this?? I think I ignored a>0 and I am really confused now