Integration by Substitution

∫5x ^{3}√(1-x^{2})*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-x__^{2})

And since it has a cube root, I did this:

1/3 *du* = ^{3}√(1-x^{2})*dx*

So I got: ∫1/3 u^{1/3} *du*

4/3 u^{4/3} + C -> 4/3 (1-x^{2})^{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-x^{2})*dx* = -5/2∫(1-x^{2})^{1/3}(-2x)*dx* = -5/2 * ((1-x^{2})^{4/3})/(4/3) + C = __(-15/8) * (1-x__^{2})^{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! :)

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Re: Integration by Substitution

Re: Integration by Substitution

Quote:

Originally Posted by

**Chaim** ∫5x ^{3}√(1-x^{2})*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-x__^{2})

And since it has a cube root, I did this:

1/3 *du* = ^{3}√(1-x^{2})*dx*

So I got: ∫1/3 u^{1/3} *du*

4/3 u^{4/3} + C -> 4/3 (1-x^{2})^{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-x^{2})*dx* = -5/2∫(1-x^{2})^{1/3}(-2x)*dx* = -5/2 * ((1-x^{2})^{4/3})/(4/3) + C = __(-15/8) * (1-x__^{2})^{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 $\displaystyle \displaystyle \begin{align*} x = \sin{(\theta)} \implies dx = \cos{(\theta)}\,d\theta \end{align*}$. Then the integral becomes

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

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

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

And remember that we originally made the substitution $\displaystyle \displaystyle \begin{align*} x = \sin{(\theta)} \end{align*}$, that means

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

which means

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

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

Re: Integration by Substitution

Hello, Chaim!

Quote:

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

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

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

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

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?

Re: Integration by Substitution

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