Is $\int_0^{\pi/2} \sin^a(\theta)\tan(\theta) d\theta $ convergent?

Question

Could someone please help me with this question? I'm not sure how i should manipulate $\sin\theta$ because of $a$.

Determine if the improper integral converges: $\int_0^{\pi/2} \sin^a(\theta)\tan(\theta) d\theta $

I've tried changing $\tan\theta$ to $ \frac{sin\theta}{\cos\theta} $ but it didn't really help me much. Maybe I should try a completely different approach...? Should I compare it to something??

Answer 1

Note that for $a \geq 0$

$$\int_0^{\pi/2-c}\sin^a(x)\tan(x) \, dx>\int_1^{\pi/2-c}\sin^a(x)\tan(x) \, dx \geq \sin^a(1)\int_1^{\pi/2-c}\tan(x) \,dx\\= \sin^a(1) [\ln(\cos(1))-\ln(\cos(\pi/2-c))],$$

and $\lim_{c \rightarrow 0}\ln(\cos(\pi/2-c))= -\infty$. Hence, the improper integral diverges.

Make a similar comparison for $a < 0$.

Answer 2

Hint:

Change variables with $x = \cos x$. Then the integral is equal to $$\lim_{t \rightarrow 0+} \int_t^1 \frac{(1-x^2)^{a/2}}{x} dx$$

Hopefully this is now conceptually clearer.