0
$\begingroup$

Please help if you can, thank you very much.

I need to prove the integrability of the following integral: $$\int_0^{\pi/2}\frac{x \sin (\tan(x))}{\cos(x)}dx$$

I tried to prove it is bounded and continuous (that's how I read I'm supposed to do it) but failed to do so, so I was hoping somebody could help out.

I also need to calculate the following integral: $$\int_0^1 \frac{\sqrt{x}}{\sqrt{1-x^6}}dx$$

I tried different approaches, but also failed. Actually, using Wolfram Alpha, it says the answer is composed of the gamma function, which seems weird as it was handed out in homework.

Thanks again

$\endgroup$
  • $\begingroup$ For the first one: are $x$, $\sin(\text{anything})$, and $\cos(\text{anything})$ bounded? Is the numerator continuous? Is the denominator continuous and nonzero on $(0,\pi/2)$? $\endgroup$ – Eric Towers Apr 20 '14 at 16:30
  • $\begingroup$ I think there's a bit more to do on the first one, because the denominator goes to $0$ at $\pi/2$, and while the value of a function at a point doesn't affect the integral, it would seem we have to be a bit more careful near $\pi/2$. $\endgroup$ – Nicholas Stull Apr 28 '14 at 13:50
1
$\begingroup$

The last integral is related to the Beta function, and hence the Gamma function you mention.

$$\int_{0}^{1}x^{1/2}(1-x^{6})^{-1/2}dx$$

Let $t=x^{6}, \;\ t^{1/6}=x, \;\ dx=1/6t^{-5/6}dt$

$$1/6\int_{0}^{1}t^{1/12}(1-t)^{-1/2}t^{-5/6}dt$$

$$1/6\int_{0}^{1}t^{-3/4}(1-t)^{-1/2}dt$$

Using the Beta function: $\beta(p,q)=\int_{0}^{1}x^{p-1}(1-x)^{q-1}dx=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$,

this means that $p=1/4, \;\ q=1/2$

Thus, we get $1/6\beta(1/4,1/2)=\frac{\Gamma(1/4)\Gamma(1/2)}{6\Gamma(3/4)}$

$$=\frac{\Gamma(1/4)\sqrt{\pi}}{6\Gamma(3/4)}$$

Wolfram gave an answer something like this?.

$\endgroup$
  • $\begingroup$ Yes, exactly. But we were not allowed to use the gamma function so I was wondering if there's a different way. $\endgroup$ – puffypuffy Apr 21 '14 at 8:30
  • $\begingroup$ I'm sorry. I even ran it through Maple and it gave the same solution in terms of Beta. So, I am thinking there may be a typo in the original problem when it was given. $\endgroup$ – Cody Apr 21 '14 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.