# Indefinite integrals

• Jan 9th 2010, 06:01 AM
mj.alawami
Indefinite integrals
Question:
$\displaystyle \int x^3.(2-x^2)^7$

$\displaystyle u=2-x^2$
$\displaystyle \int x^3 .u^7$
$\displaystyle du=-2x$
I dont know what to do next.
• Jan 9th 2010, 06:10 AM
Prove It
Quote:

Originally Posted by mj.alawami
Question:
$\displaystyle \int x^3.(2-x^2)^7$

$\displaystyle u=2-x^2$
$\displaystyle \int x^3 .u^7$
$\displaystyle du=-2x$
I dont know what to do next.

For this you should use integration by parts.

Remember that $\displaystyle \int{u\,dv} = uv - \int{v\,du}$.

In this case, let $\displaystyle u = x^2$ so that $\displaystyle du = 2x$

Let $\displaystyle dv = x(2 - x^2)^7$ so that $\displaystyle v = \int{x(2 - x^2)^7\,dx}$.

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

$\displaystyle = -\frac{1}{2}\cdot\frac{1}{8}(2 - x^2)^8$

$\displaystyle = -\frac{1}{16}(2 - x^2)^8$.

So $\displaystyle \int{x^3(2 - x^2)^7\,dx} = -\frac{1}{16}x^2(2 - x^2)^8 - \int{-\frac{1}{8}x(2 - x^2)^8\,dx}$

$\displaystyle = -\frac{1}{16}x^2(2 - x^2)^8 - \frac{1}{16}\int{-2x(2 - x^2)^8\,dx}$

$\displaystyle = -\frac{1}{16}x^2(2 - x^2)^8 - \frac{1}{16}\cdot\frac{1}{9}(2 - x^2)^9 + C$

$\displaystyle = -\frac{1}{16}x^2(2 - x^2)^8 - \frac{1}{144}(2 - x^2)^9 + C$.
• Jan 9th 2010, 06:21 AM
chisigma
The substitution $\displaystyle u=2-x^{2}$ is itself correct, but the complete procedure is...

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

$\displaystyle dx= -\frac{du}{2x}= -\frac{du}{2\cdot (2-u)^{\frac{1}{2}}}$ (2)

... so that the integral becomes...

$\displaystyle - \frac{1}{2}\cdot \int u^{7} \cdot (2-u)\cdot du$ (3)

... that really seemes more easy than the original one...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jan 9th 2010, 07:05 AM
HallsofIvy
Quote:

Originally Posted by mj.alawami
Question:
$\displaystyle \int x^3.(2-x^2)^7$

$\displaystyle u=2-x^2$
$\displaystyle \int x^3 .u^7$
$\displaystyle du=-2x$
I dont know what to do next.
Again, "du= -2x" is not correct. du= -2xdx so $\displaystyle -\frac{1}{2}du= xdx$. You need to learn to write that correctly!
Rewrite $\displaystyle \int x^2(2-x^2)^7 dx$ as $\displaystyle \int x^2(2- x^2)^7(xdx)$ Now $\displaystyle (2- x^2)^7$ becomes $\displaystyle u^7$, $\displaystyle x dx$ becomes $\displaystyle -\frac{1}{2} du$ and $\displaystyle x^2$ becomes 2- u.
$\displaystyle \int x^3(2- x^2)^7 dx$ becomes $\displaystyle -\frac{1}{2}\int (2-u)u^7 du= -\frac{1}{2}\int (2u^7- u^8) du$