MHB Indefinite Integral using Trig Identity i'm confused

[Pindrought]

Okay so I'm working on this problem

$$\int \frac{x^2}{\sqrt{4 - x^2}} \, dx$$

I do a substitution and set
$$x={\sqrt{4}}sinu$$

I get to this step fine

$$\int 4sin(u)^2$$

I know that u = arcsin(x/2)

so I don't see why I can't just substitute in u into sin(u)?

I tried this and I got
$$\int 4 * arcsin(sin(x/2))^2$$

which worked out to
$$\int 4 * \frac{x^2}{4}$$

which gave me
$$\int x^2$$

which would just mean the answer is
$$\frac{x^3}{3}$$

But looking at the mathhelpboards solver, this is wrong. Can anyone help me figure out what I am not understanding? Thanks a lot for taking the time to read.

 

[ThePerfectHacker]

Hint: $2\sin^2 u = 1 - \cos 2u$.

 

[Chris L T521]

Okay so I'm working on this problem

$$\int \frac{x^2}{\sqrt{4 - x^2}} \, dx$$

I do a substitution and set
$$x={\sqrt{4}}sinu$$

I get to this step fine

$$\int 4sin(u)^2$$

I know that u = arcsin(x/2)

so I don't see why I can't just substitute in u into sin(u)?

I tried this and I got
$$\int 4 * arcsin(sin(x/2))^2$$

which worked out to
$$\int 4 * \frac{x^2}{4}$$

which gave me
$$\int x^2$$

which would just mean the answer is
$$\frac{x^3}{3}$$

But looking at the mathhelpboards solver, this is wrong. Can anyone help me figure out what I am not understanding? Thanks a lot for taking the time to read.

The differentials you didn't include in the integrals give us an idea why your approach isn't correct.

What you start with is

\[\int \frac{x^2}{\sqrt{4-x^2}}\!\color{red}{\,dx}\]

We then make the trig substitution $x=2\sin u\implies \color{red}{\,dx} = 2\cos u \color{blue}{\,du}$

Therefore,

\[\begin{aligned} \int \frac{x^2}{\sqrt{4-x^2}}\!\color{red}{\,dx} \xrightarrow{x=2\sin u}{} &\phantom{=}\int\frac{4\sin^2u}{2\cos u}\cdot 2\cos u\color{blue}{\,du} \\ &= \int 4\sin^2u\color{blue}{\,du} \end{aligned}\]

If we proceed with how you wanted to do things, if $x=2\sin u \implies u = \arcsin(x/2)$, then

\[\int 4\sin^2u\,du = \int 4\sin^2(\arcsin(x/2))\,du = \int x^2\color{blue}{\,du}\]

We can't say this is $\dfrac{x^3}{3}+C$ because the variable we're integrating with respect to here is $u$, not $x$! If you want to get $\,dx$ instead of $\,du$ in the integral, you are undoing what you just did with your substitution.

Instead, from $\displaystyle \int 4\sin^2u\,du$, you want to apply the trig identity $\sin^2\theta = \dfrac{1-\cos(2\theta)}{2}$ to get
\[\int 4\sin^2u\,du = \int 2-2\cos(2u)\,du\]

Can you take things from here? (Smile)

 

[Pindrought]

The differentials you didn't include in the integrals give us an idea why your approach isn't correct.

What you start with is

\[\int \frac{x^2}{\sqrt{4-x^2}}\!\color{red}{\,dx}\]

We then make the trig substitution $x=2\sin u\implies \color{red}{\,dx} = 2\cos u \color{blue}{\,du}$

Therefore,

\[\begin{aligned} \int \frac{x^2}{\sqrt{4-x^2}}\!\color{red}{\,dx} \xrightarrow{x=2\sin u}{} &\phantom{=}\int\frac{4\sin^2u}{2\cos u}\cdot 2\cos u\color{blue}{\,du} \\ &= \int 4\sin^2u\color{blue}{\,du} \end{aligned}\]

If we proceed with how you wanted to do things, if $x=2\sin u \implies u = \arcsin(x/2)$, then

\[\int 4\sin^2u\,du = \int 4\sin^2(\arcsin(x/2))\,du = \int x^2\color{blue}{\,du}\]

We can't say this is $\dfrac{x^3}{3}+C$ because the variable we're integrating with respect to here is $u$, not $x$! If you want to get $\,dx$ instead of $\,du$ in the integral, you are undoing what you just did with your substitution.

Instead, from $\displaystyle \int 4\sin^2u\,du$, you want to apply the trig identity $\sin^2\theta = \dfrac{1-\cos(2\theta)}{2}$ to get
\[\int 4\sin^2u\,du = \int 2-2\cos(2u)\,du\]

Can you take things from here? (Smile)

Thanks a lot Chris L T521, you really helped me understand what I was doing wrong.