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

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??

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 $a < 0$.

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.

We start with the definition of the Beta function: $$B(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ Where $$\Gamma(\cdot)$$ denotes the Gamma function. If $$a$$ and $$b$$ are complex numbers, this identity holds (AKA the Beta function converges) for $$\operatorname{Re}(a)>0,\qquad \operatorname{Re}(b)>0$$ If $$a$$ and $$b$$ are real, $$a>0,\qquad b>0$$ Again if $$a$$ and/or $$b$$ do not satisfy the above inequalities, the $$B(a,b)$$ does not converge.

Now consider the integral $$I(a,b)=\int_0^{\pi/2}\sin^ax\ \cos^bx\ dx$$ The substitution $$t=\sin^2x$$ gives $$I(a,b)=\frac12\int_0^1t^{\frac{a-1}2}(1-t)^{\frac{b-1}2}dt$$ Which in turn gives $$I(a,b)=\frac12B\bigg(\frac{a+1}2,\frac{b+1}2\bigg)$$ As you may have noticed, your integral can be written as $$I(a+1,-1)$$. Since $$-1\not >0$$, we know that your integral diverges.