# Indefinite Integral using Trig Identity i'm confused

#### Pindrought

##### New member
Okay so i'm working on this problem

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

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

I get to this step fine

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

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
$$\displaystyle \int 4 * arcsin(sin(x/2))^2$$

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

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

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

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

##### Well-known member
Hint: $2\sin^2 u = 1 - \cos 2u$.

#### Chris L T521

##### Well-known member
Staff member
Okay so i'm working on this problem

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

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

I get to this step fine

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

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
$$\displaystyle \int 4 * arcsin(sin(x/2))^2$$

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

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

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

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.

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

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? #### Pindrought

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

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

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