Does the improper integral $ \int_0^ \frac{ \pi }{2} \sin ^{a}\theta \tan\theta d\theta$ converge?

Question

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

Answer 1

It does not converge. Consider that: $\sin(x) \geq \dfrac{2x}{\pi}$ for $0\leq x\leq\dfrac{\pi}{2}$.

$$\int_0^{\pi/2} \sin^{a}(x)\tan(x)\, \mathrm{d}x \geq \int_0^{\pi/2} \left(\dfrac{2x}{\pi}\right)^{a}\tan(x)\, \mathrm{d}x$$

If the second integral diverges then the first one must diverge also. It must be that second one diverges.

We can construct another inequality to show divergence.

$$\int_0^{\pi/2} \left(\dfrac{2x}{\pi}\right)^{a}\tan(x)\, \mathrm{d}x \geq C\int_1^{\pi/2} x^a\tan(x)\, \mathrm{d}x \geq C\int_1^{\pi/2} \tan(x)\, \mathrm{d}x$$

This holds for some $C$. The last integral diverges so all of the integrals must diverge.

Answer 2

Letting $u=\cos(x)$, we get $$ \begin{align} \int_0^{\pi/2}\sin^a(x)\tan(x)\,\mathrm{d}x &=-\int_0^{\pi/2}\sin^a(x)\frac{\mathrm{d}\cos(x)}{\cos(x)}\\ &=\int_0^1(1-u^2)^{a/2}\frac{\mathrm{d}u}{u}\\ &\ge\int_0^{1/2}(3/4)^{a/2}\frac{\mathrm{d}u}{u} \end{align} $$