Integration by Substitution

**Chaim** · February 5th 2013, 07:46 PM

∫5x ³√(1-x²)dx

So I'm still a beginner with integrating by substituting.
What I know is that you use 'u' from one of the thing in the equation to substitute.

Though, I'm not entirely sure where 'u' should be coming from.
So I did u = (1-x²)
And since it has a cube root, I did this:
1/3 du = ³√(1-x²)dx
So I got: ∫1/3 u^1/3 du
4/3 u^4/3 + C -> 4/3 (1-x²)³ + C
But I'm not sure if this is entirely correct after:

Someone else showed me how to do the steps, which was this:
∫5x ³√(1-x²)dx = -5/2∫(1-x²)^1/3(-2x)dx = -5/2 * ((1-x²)^4/3)/(4/3) + C = (-15/8) * (1-x²)^4/3 + C
This was the correct answer in the book.
But I'm a bit confused on how the person got the -5/2, and the (-2x)
Can someone explain to me that part?
Thanks!

**ibdutt** · February 5th 2013, 08:46 PM

**Prove It** · February 5th 2013, 08:55 PM

Originally Posted by Chaim

∫5x ³√(1-x²)dx

So I'm still a beginner with integrating by substituting.
What I know is that you use 'u' from one of the thing in the equation to substitute.

Though, I'm not entirely sure where 'u' should be coming from.
So I did u = (1-x²)
And since it has a cube root, I did this:
1/3 du = ³√(1-x²)dx
So I got: ∫1/3 u^1/3 du
4/3 u^4/3 + C -> 4/3 (1-x²)³ + C
But I'm not sure if this is entirely correct after:

Someone else showed me how to do the steps, which was this:
∫5x ³√(1-x²)dx = -5/2∫(1-x²)^1/3(-2x)dx = -5/2 * ((1-x²)^4/3)/(4/3) + C = (-15/8) * (1-x²)^4/3 + C
This was the correct answer in the book.
But I'm a bit confused on how the person got the -5/2, and the (-2x)
Can someone explain to me that part?
Thanks!

The better substitution to use here is $\displaystyle \begin{align*} x = \sin{(\theta)} \implies dx = \cos{(\theta)}\,d\theta \end{align*}$ . Then the integral becomes

$\displaystyle \begin{align*} \int{\frac{5x^3}{\sqrt{1 - x^2}}\,dx} &= \int{\frac{5\sin^3{(\theta)}}{\sqrt{1 - \sin^2{(\theta)}}}\, \cos{(\theta)}\,d\theta} \\ &= \int{\frac{5\sin^3{(\theta)}\cos{(\theta)}}{\sqrt{ \cos^2{(\theta)}}}\,d\theta} \\ &= \int{\frac{5\sin^3{(\theta)}\cos{(\theta)}}{\cos{( \theta)}}\,d\theta} \\ &= \int{5\sin^3{(\theta)}\,d\theta} \\ &= \int{5\sin{(\theta)}\sin^2{(\theta)}\,d\theta} \\ &= -5\int{-\sin{(\theta)}\left[ 1 - \cos^2{(\theta)} \right] d\theta} \end{align*}$

Now make the substitution $\displaystyle \begin{align*} u = \cos{(\theta)} \implies du = -\sin{(\theta)}\,d\theta \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} -5\int{-\sin{(\theta)}\left[ 1 - \cos^2{(\theta)} \right] d\theta} &= -5\int{ 1 - u^2 \,du} \\ &= -5 \left( u - \frac{u^3}{3} \right) + C \\ &= -5 \left[ \cos{(\theta)} - \frac{\cos^3{(\theta)}}{3} \right] + C \end{align*}$

And remember that we originally made the substitution $\displaystyle \begin{align*} x = \sin{(\theta)} \end{align*}$ , that means

$\displaystyle \begin{align*} x^2 &= \sin^2{(\theta)} \\ 1 - x^2 &= 1 - \sin^2{(\theta)} \\ 1 - x^2 &= \cos^2{(\theta)} \\ \sqrt{1 - x^2} &= \cos{(\theta)} \end{align*}$

which means

$\displaystyle \begin{align*} -5 \left[ \cos{(\theta)} - \frac{\cos^3{(\theta)}}{3} \right] +C &= -5 \left[ \sqrt{1 - x^2} - \frac{\left( \sqrt{1 -x^2} \right) ^3 }{3} \right] + C \end{align*}$

So finally we have $\displaystyle \begin{align*} \int{ \frac{5x^3}{\sqrt{1 - x^2}}\,dx } = -5 \left[ \sqrt{1 - x^2} - \frac{\left( \sqrt{1 - x^2} \right) ^3}{3} \right] + C \end{align*}$

**Soroban** · February 5th 2013, 09:59 PM

Hello, Chaim!

$\int 5x\sqrt[3]{1-x^2}\,dx$

We have: . $5\int(1-x^2)^{\frac{1}{3}}(x\,dx)$

Let $u \,=\,1-x^2 \quad\Rightarrow\quad du \,=\,\text{-}2x\,dx \quad\Rightarrow\quad x\,dx \,=\,\text{-}\tfrac{1}{2}du$

Substitute: . $5\int u^{\frac{1}{3}}(\text{-}\tfrac{1}{2}\,du) \;=\; \text{-}\tfrac{5}{2}\int u^{\frac{1}{3}}\,du \;=\; \text{-}\tfrac{15}{8}u^{\frac{4}{3}}+C$

Back-substitute: . $\text{-}\tfrac{15}{8}(1-x^2)^{\frac{4}{3}} + C$

**Chaim** · February 6th 2013, 08:59 AM

Thanks for responding guys!
I understand how u and du is coming from now.

I'm a bit confused on one thing.
The xdx

Just wondering, how come xdx = -1/2?

**HallsofIvy** · February 6th 2013, 10:08 AM

It isn't and Soroban didn't say it was. From $u= 1- x^2$ , we get $\frac{du}{dx}= -2x$ so that $du= -2x dx$ . Divide both sides by -2 to get $-\frac{1}{2}du= xdx$ .

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