# Math Help - Integral Refresher Question

1. ## Integral Refresher Question

Hi, I'm having a little bit of trouble with a integration question, here is the text from the question.

Find the antiderivitive of the given functions:

$F(x) = x(1-x^2)^7$

Basically I just have no idea how to expand $(1 - x^2)^7$

2. Originally Posted by Crell
Hi, I'm having a little bit of trouble with a integration question, here is the text from the question.

Find the antiderivitive of the given functions:

$F(x) = x(1-x^2)^7$

Basically I just have no idea how to expand $(1 - x^2)^7$
Dont expand!

Use substition.

u=1-x^2

3. Well, you could expand it using the "binomial theorem": $(a+ b)^n= \sum_{i=0}^n \begin{pmatrix}n \\ i\end{pmatrix}a^ib^{n- i}$
For $(1- x^2)^7= 1- 7x^2+ 21x^4+ \cdot\cdot\cdot- 7x^{12}+ x^{14}$

But you don't want to do that. Instead, make the substituition $u= 1- x^2$ so that $(1- x^2)^7= u^7$ and $du= -2 xdx$.

4. Originally Posted by HallsofIvy
$du= -2 xdx$.
Thank you much, this is the part I was missing.

$

\int x(1-x^2)^7dx

$

Aside:

Let $u = 1-x^2$

Derive this equation

$\frac{du}{dx} = -2x$

Solve for $dx$

$dx = \frac{du}{-2x}$

Manipulate the first equation to make it appear like $-2x(u)^7dx$ so that when you substitute dx, the x is eliminated.

Therefore:

$-\frac{1}{2} \int -2x(u)^7dx$

Substitute $dx = \frac{du}{-2x}$ into first equation
$-\frac{1}{2} \int \frac {-2x(u)^7du}{-2x}$

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

Integrate, sub in u and simplfy.

$- \frac{1}{2} [\frac {(u)^8}{8}]$

$-\frac{1}{2} [\frac {(1-x^2)^8}{8}]$

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

5. Did you differentiate that answer to cheek it for yourself?

6. Have now, thanks.

7. This is a perfect place to use my favourite rule of integration:

$\displaystyle\int{f'(x)f(x)^n} dx=\frac{1}{n+1}f(x)^{n+1}+c$

So let $f(x)=(1-x^2)$
$f'(x)=-2x$
$n=7$

$f'(x)f(x)^n= -2x(1-x^2)^7$
You'll notice that it isn't quite a match to what you are trying to integrate. However, if you divide by $-2$, it is exactly what you are trying to integrate. Therefore, we can apply the rule as normal, but we must remember to divide our final answer by -2.

$\displaystyle\int x(1-x^2)^7 dx$

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

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

I think in this case, this method is much nicer than substitution - if you spot the rule, it will take a few lines of working to obtain the full solution with no messy work required at all.

8. In speaking of the binomial theorem, you can find the anti-derivative
without performing an expansion, but in somewhat undesirable forms:

$\displaystyle \sum_{k=0}^{\infty}\frac{(-1)^kx^{2k+2}}{2k+2}\binom{7}{k}$ or $\displaystyle \sum_{k=0}^{7}\frac{(-1)^k(x)^{2k+2}}{2k+2}\binom{7}{k}$.

Plus the constant of integration. The u-sub does the job far better.

Spoiler:
1.

$\displaystyle x(1-x^2)^7 = x\left(\sum_{k=0}^{\infty}\binom{7}{k}(-1)^kx^{2k}\right) = \sum_{k=0}^{\infty}\binom{7}{k}(-1)^{k}x^{2k+1}.$

\begin{aligned}\displaystyle \therefore \int x(1-x^2)^7\;{dx} & = \int \left(\sum_{k=0}^{\infty}\binom{7}{k}(-1)^{k}x^{2k+1}\right)\;{dx} \\& = \sum_{k=0}^{\infty} \left(\int \binom{7}{k}(-1)^{k}x^{2k+1}\;{dx}\right) \\& = \sum_{k=0}^{\infty}\frac{(-1)^kx^{2k+2}}{2k+2}\binom{7}{k}+C.\end{aligned}

2.

\displaystyle \begin{aligned}x(1-x^2)^7 & = x\left(\sum_{k=0}^{7}\binom{7}{k}(-1)^kx^{2k}\right) \\& = \sum_{k=0}^{7}\binom{7}{k}(-1)^kx^{2k+1}.\end{aligned}

\displaystyle \begin{aligned} \therefore \int x(1-x^2)^7\;{dx} & = \int \left(\sum_{k=0}^{7}\binom{7}{k}(-1)^kx^{2k+1}\right)\;{dx} \\& = \sum_{k=0}^{7} \left(\int \binom{7}{k}(-1)^kx^{2k+1}\;{dx}\right) \\& = \sum_{k=0}^{7}\frac{(-1)^k(x)^{2k+2}}{2k+2}\binom{7}{k}+C.\end{aligned}