Cfrgtkky

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

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

I've tried changing $tantheta$ to $ frac{sintheta}{costheta} $ 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
$$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}dt$$
Which in turn gives
$$I(a,b)=frac12Bbigg(frac{a+1}2,frac{b+1}2bigg)$$
As you may have noticed, your integral can be written as $I(a+1,-1)$. Since $-1not >0$, we know that your integral diverges.