MHB How Do You Evaluate ∫x²√(1-x²) dx Using Trig Substitution?

[karush]

W8.3.6 evaluate
$$\int {x}^{2}\sqrt{1-{x}^{2 }} \ dx
= \arcsin\left({x}\right)/8—\sin\left({4\arcsin\left({x}\right)}\right)/32 + C $$
This is from an exercise on trig substitutions so
$$x=\sin\left({x}\right)
\text{ so }
\int\sin^2 \left({x}\right)\sqrt{1-\sin^2 \left({x}\right)}\ dx
\implies \int\sin^2 \left({x}\right) \cos\left({x}\right) \ dx
$$
No example to follow on this so seeing if this is a good start

 

[MarkFL]

I would let:

$$x=\sin(\theta)\implies dx=\cos(\theta)\,d\theta$$

And the integral becomes:

$$I=\int \sin^2(\theta)\cos^2(\theta)\,d\theta$$

At this point I would look at the double-angle identity for sine...:)

 

[karush]

$\sin\left({2x}\right)=2\sin\left({x}\right)\cos\left({x}\right)$
Then
$I=2\int\sin^2 \left({2x}\right)\ dx $
?

 

[MarkFL]

You are correct when you state:

$$\sin(2u)=2\sin(u)\cos(u)$$

However, this then means:

$$\sin(u)\cos(u)=\frac{\sin(2u)}{2}$$

And so the integral becomes:

$$I=\frac{1}{4}\int \sin^2(2\theta)\,d\theta$$

Now a double-angle identity for cosine can be used...the form which relates the cosine of a double-angle to a function of the square of the sine function...:)

 

[karush]

Double angle of cosine can be used?

$$\cos\left({2\theta}\right)=1-2\sin^2 \left({\theta}\right)$$

 

[Greg]

$$\sin^2(x)=1-\cos^2(x)$$

$$2\sin^2(x)=1-\cos^2(x)+\sin^2(x)$$

$$2\sin^2(x)=1-\cos(2x)$$

$$\sin^2(x)=\dfrac{1-\cos(2x)}{2}$$

 

[karush]

So then we would have

$$I=\frac{1}{4}\int\frac{1}{2} \ dx
- \frac{1}{4 }\int\frac{\cos\left({2x}\right)}{2} \ dx
\implies
\frac{x}{8} - \frac{\sin\left({2x}\right)}{16} + C
$$
This doesn't look like the answer!

 

[Prove It]

So then we would have

$$I=\frac{1}{4}\int\frac{1}{2} \ dx
- \frac{1}{4 }\int\frac{\cos\left({2x}\right)}{2} \ dx
\implies
\frac{x}{8} - \frac{\sin\left({2x}\right)}{16} + C
$$
This doesn't look like the answer!

You have forgotten that you converted to a function of $\displaystyle \begin{align*} \theta \end{align*}$.

So from $\displaystyle \begin{align*} \frac{\theta}{8} - \frac{\sin{(2\,\theta )}}{16} + C \end{align*}$ and recalling your original substitution of $\displaystyle \begin{align*} x = \sin{(\theta)} \end{align*}$ what is the answer?

 

[MarkFL]

Double angle of cosine can be used?

$$\cos\left({2\theta}\right)=1-2\sin^2 \left({\theta}\right)$$

Yes, from this you get (I will use $u$ instead of $\theta$ since we are already using $\theta$):

$$\sin^2(u)=\frac{1-\cos(2u)}{2}$$

And so your integral then becomes:

$$I=\frac{1}{8}\int 1-\cos(4\theta)\,d\theta$$

 

[karush]

earlier
$x=\sin\left({\theta}\right)$
so
$\theta = \arcsin\left({x}\right)$
Then back substittute
$\displaystyle \frac{\arcsin\left({x}\right)}{8} - \frac{\sin\left({4\arcsin\left({x}\right)}\right)}{32}+C$