# Integration by Substitution

• February 5th 2013, 08:46 PM
Chaim
Integration by Substitution
∫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 u1/3 du
4/3 u4/3 + 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! :)
• February 5th 2013, 09:46 PM
ibdutt
Re: Integration by Substitution
• February 5th 2013, 09:55 PM
Prove It
Re: Integration by Substitution
Quote:

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 u1/3 du
4/3 u4/3 + 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*}
• February 5th 2013, 10:59 PM
Soroban
Re: Integration by Substitution
Hello, Chaim!

Quote:

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

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: . $\text{-}\tfrac{15}{8}(1-x^2)^{\frac{4}{3}} + C$
• February 6th 2013, 09:59 AM
Chaim
Re: Integration by Substitution
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?
• February 6th 2013, 11:08 AM
HallsofIvy
Re: Integration by Substitution
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 $-\frac{1}{2}du= xdx$.