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Indefinite Integral using Trig Identity i'm confused

Pindrought

New member
Mar 3, 2014
15
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
Jan 26, 2012
236
Hint: $2\sin^2 u = 1 - \cos 2u$.
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
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.

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

New member
Mar 3, 2014
15
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.