Evaluating integrals

**khch313** · July 13th 2010, 08:21 PM

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!!

**Prove It** · July 13th 2010, 08:59 PM

For the second

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

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

Then the integral becomes

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

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

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

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

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

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

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

So

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

$= \frac{1}{2}\arcsin{x} - \frac{1}{2}x\sqrt{1 - x^2} + C$ .

**khch313** · July 13th 2010, 09:19 PM

Thank you!

**tom@ballooncalculus** · July 14th 2010, 09:56 AM

Just in case a picture helps with the other...

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

Spoiler:

Do long division to split up the improper fraction,

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

Spoiler:

________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

**khch313** · July 14th 2010, 10:52 AM

Originally Posted by tom@ballooncalculus

Just in case a picture helps with the other...

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

Spoiler:

Do long division to split up the improper fraction,

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

Spoiler:

________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

Thank you for this. I have never seen this kind of diagram before

**Prove It** · July 14th 2010, 06:13 PM

Or without a picture...

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

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

So the integral becomes

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

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

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

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

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

Thread: Evaluating integrals

LinkBack

Thread Tools

Display

Evaluating integrals

Similar Math Help Forum Discussions

Evaluating the Integrals

evaluating integrals

Some help with evaluating integrals

Evaluating Integrals

Evaluating Integrals

Search Tags