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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

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    $\begingroup$ The details are not right, but substitution will lead to an answer. You should not, however, depend on finding an explicit antiderivative. My preference would be to deal directly with the "badness" at $\pi/2$. Hint: There is a $k$ such that $\sin^a \theta\gt \frac{1}{2}$ if $k\le \theta\lt \pi/2$. $\endgroup$ – André Nicolas May 8 '14 at 18:11
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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.

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

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