# Thread: similar to gamma function

1. ## similar to gamma function

ive got this question in for tomorrow and i have no idea whatsoever how to attempt it

evaluate the following intergral. your final answer should not contain the gamma function.

maths 6.JPG

thank you

2. Since you can't use gamma, you could try rewriting and making a substitution.

$\displaystyle (1-x^{3})^{\frac{5}{2}}\cdot{x^{\frac{7}{2}}}$

$\displaystyle =(1-x^{3})^{2}(1-x^{3})^{\frac{1}{2}}x^{\frac{7}{2}}$

Let $\displaystyle u=\sqrt{1-x^{3}}, \;\ du=\frac{-3x^{2}}{2\sqrt{1-x^{3}}}$

Making the subs gives:

$\displaystyle \frac{-2}{3}\int{u^{6}\sqrt{1-u^{2}}}du$

Now, let $\displaystyle u=sin{\theta}, \;\ du=cos{\theta}d{\theta}$

Make the subs and you get:

$\displaystyle \frac{-2}{3}\int{sin^{6}{\theta}cos^{2}{\theta}}d{\theta}$

$\displaystyle =\frac{-2}{3}\int[sin^{6}{\theta}-sin^{8}{\theta}]d{\theta}$

Now, you can use the reduction formulas:

$\displaystyle \int{sin^{n}{\theta}}d{\theta}=\frac{sin^{n-1}{\theta}cos{\theta}}{n}+\frac{n-1}{n}\int{sin^{n-2}}d{\theta}$

Just a thought. It ain't pretty no matter how you look at it.

3. This can be achieved with the Beta Integral.

$\displaystyle \int_0^1 (1-x^3)^{5/2} x^{7/2} dx$
We can write this as,
$\displaystyle \frac{1}{3} \int_0^1 (1-x^3)^{5/2} \cdot x^{3/2} \cdot (3x^2) \ dx$

Let $\displaystyle t=x^3 \implies t'=3x^2$ thus,

$\displaystyle \frac{1}{3}\int_0^1 (1-t)^{5/2} \cdot t^{1/2} dt$

Use the Beta Integral and get,
$\displaystyle \bold{B}\left(\frac{7}{2} , \frac{3}{2} \right)$

Converte to Gamma Function,

$\displaystyle \frac{\Gamma \left( \frac{7}{2} \right) \cdot \Gamma \left( \frac{3}{2} \right) }{\Gamma (5)}$

But,
$\displaystyle \Gamma \left( \frac{1}{2} \right) = \sqrt{\pi}$

Thus, using reduction identity for Gamma functions,
$\displaystyle \Gamma \left( \frac{7}{2} \right) = \frac{15\sqrt{\pi}}{8}$

And,
$\displaystyle \Gamma \left( \frac{3}{2} \right) = \frac{\sqrt{\pi}}{2}$

And,
$\displaystyle \Gamma (5) = 4!=24$

Thus,
$\displaystyle \mbox{ answer } = \frac{15\pi}{384}$

4. Galactus' approach is not completely way off. He arrives at the integral,
$\displaystyle \int_0^{\pi/2} \sin ^{\alpha} x \cos^{\beta} x \ dx$

Which is discussed here.
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