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!!
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
$\displaystyle \frac{1}{2}\theta - \frac{1}{4}\sin{2\theta} + C = \frac{1}{2}\theta - \frac{1}{2}\sin{\theta}\cos{\theta} + C$
$\displaystyle = \frac{1}{2}\arcsin{x} - \frac{1}{2}x\sqrt{1 - x^2} + C$.
Just in case a picture helps with the other...
... 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!
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 $\displaystyle x = 1, u = 1$.