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??
 A: 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 $a < 0$.
A: 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.
A: 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.
