Hello! I have done all of my math homework but I'm so stuck on these couple problems.'

0∫1 [ 1/(1+∛x) ]dx (integral between 0 and 1)

and

∫ [ x²/√(1-x²) ]dx

Thank you!!

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- Jul 13th 2010, 07:21 PMkhch313Evaluating integrals
Hello! I have done all of my math homework but I'm so stuck on these couple problems.'

0∫1 [ 1/(1+∛x) ]dx (integral between 0 and 1)

and

∫ [ x²/√(1-x²) ]dx

Thank you!! - Jul 13th 2010, 07:59 PMProve It
For the second

$\displaystyle \int{\frac{x^2}{\sqrt{1-x^2}}\,dx}$

make the substitution $\displaystyle x = \sin{\theta}$ so that $\displaystyle dx = \cos{\theta}\,d\theta$.

Then the integral becomes

$\displaystyle \int{\frac{\sin^2{\theta}}{\sqrt{1 - \sin^2{\theta}}}\,\cos{\theta}\,d\theta}$

$\displaystyle = \int{\frac{\cos{\theta}\sin^2{\theta}}{\sqrt{\cos^ 2{\theta}}}\,d\theta}$

$\displaystyle = \int{\frac{\cos{\theta}\sin^2{\theta}}{\cos{\theta }}\,d\theta}$

$\displaystyle = \int{\sin^2{\theta}\,d\theta}$

$\displaystyle = \int{\frac{1}{2} - \frac{1}{2}\cos{2\theta}\,d\theta}$

$\displaystyle = \frac{1}{2}\theta - \frac{1}{4}\sin{2\theta} + C$.

Now remembering that $\displaystyle x = \sin{\theta}$ that means $\displaystyle \theta = \arcsin{x}$ and $\displaystyle \cos{\theta} = \sqrt{1 - \sin^2{\theta}} = \sqrt{1 - x^2}$.

So

$\displaystyle \frac{1}{2}\theta - \frac{1}{4}\sin{2\theta} + C = \frac{1}{2}\theta - \frac{1}{2}\sin{\theta}\cos{\theta} + C$

$\displaystyle = \frac{1}{2}\arcsin{x} - \frac{1}{2}x\sqrt{1 - x^2} + C$. - Jul 13th 2010, 08:19 PMkhch313
Thank you!

- Jul 14th 2010, 08:56 AMtom@ballooncalculus
Just in case a picture helps with the other...

http://www.ballooncalculus.org/asy/internal/three.png

... where... (key in spoiler)

__Spoiler__:

Do long division to split up the improper fraction,

$\displaystyle \displaystyle{\frac{u^2}{1 + u}\ =\ \frac{u(1 + u) - (1 + u) + 1}{1 + u}}$

__Spoiler__:

________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote! - Jul 14th 2010, 09:52 AMkhch313
- Jul 14th 2010, 05:13 PMProve It
Or without a picture...

$\displaystyle \int{\frac{1}{1 + \sqrt[3]{x}}\,dx}$.

Let $\displaystyle u = \sqrt[3]{x}$ so that $\displaystyle x = u^3$ and $\displaystyle dx = 3u^2\,du$.

So the integral becomes

$\displaystyle \int{\left(\frac{1}{1 + u}\right)3u^2\,du}$

$\displaystyle = 3\int{\frac{u^2}{1 + u}\,du}$

$\displaystyle = 3\int{u - 1 + \frac{1}{1 + u}\,du}$

$\displaystyle = 3\left(\frac{1}{2}u^2 - u + \ln{|1+u|} + C\right)$.

Now substitute your terminals, remembering that if $\displaystyle x = 0, u = 0$ and if $\displaystyle x = 1, u = 1$.