Indefinite integrals

**mj.alawami** · January 9th 2010, 06:01 AM

Question:
$\int x^3.(2-x^2)^7$

$u=2-x^2$
$\int x^3 .u^7$
$du=-2x$
I dont know what to do next.
Please answer in steps ,Thank you

**Prove It** · January 9th 2010, 06:10 AM

Originally Posted by mj.alawami

Question:
$\int x^3.(2-x^2)^7$

$u=2-x^2$
$\int x^3 .u^7$
$du=-2x$
I dont know what to do next.
Please answer in steps ,Thank you

For this you should use integration by parts.

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

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

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

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

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

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

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

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

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

$= -\frac{1}{16}x^2(2 - x^2)^8 - \frac{1}{144}(2 - x^2)^9 + C$ .

**chisigma** · January 9th 2010, 06:21 AM

The substitution $u=2-x^{2}$ is itself correct, but the complete procedure is...

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

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

... so that the integral becomes...

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

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

Kind regards

$\chi$ $\sigma$

**HallsofIvy** · January 9th 2010, 07:05 AM

Originally Posted by mj.alawami

Question:
$\int x^3.(2-x^2)^7$

$u=2-x^2$
$\int x^3 .u^7$
$du=-2x$
I dont know what to do next.
Please answer in steps ,Thank you

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

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