How can trigonometric substitution simplify integration?

[karush]

$\int{x}^{2}\sqrt{1-{x}^{2}}dx$
$u=\sec\left({x}\right)\ du= \frac{\sin\left({x}\right)}{{\cos\left({x}\right)}^{2}}dx$

I pursued this but got lost, maybe I don't need a trig subst.

 

[topsquark]

$\int{x}^{2}\sqrt{1-{x}^{2}}dx$
$u=\sec\left({x}\right)\ du= \frac{\sin\left({x}\right)}{{\cos\left({x}\right)}^{2}}dx$

I pursued this but got lost, maybe I don't need a trig subst.

Try u = sin(theta) or x = cos(theta).

-Dan

 

[karush]

$x=\sin\left({\theta}\right)\ dx=\cos\left({\theta}\right)$

$\int\sin^2\left({\theta}\right)\cos\left({\theta}\right)d\theta$

 

[topsquark]

$x=\sin\left({\theta}\right)\ dx=\cos\left({\theta}\right)$

$\int\sin^2\left({\theta}\right)\cos\left({\theta}\right)d\theta$

That's the one! Can you finish it from here?

-Dan

 

[karush]

$u=\sin\left({\theta}\right)
\ du=\cos\left({\theta}\right)\ d\theta$
$\int{u}^{2}\ du$

This isn't heading towards the TI answer

 

[MarkFL]

If you let:

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

Then your integral becomes:

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

At this point, I would then rewrite the integral as:

$$I=\frac{1}{8}\int \sin^2(2\theta)\,d(2\theta)$$

Now, use a double-angle identity for cosine to proceed...:D

 

[karush]

Where does the 1/8 come from?

 

[MarkFL]

Where does the 1/8 come from?

It comes from:

i) $$\sin^2(2u)=4\sin^2(u)\cos^2(u)$$

ii) We want the differential to be $$d(2\theta)$$

Does that make sense?

 

[karush]

Yes but have never seen $dx$ look like $d(2\theta)$

 

[MarkFL]

Yes but have never seen $dx$ look like $d(2\theta)$

It allows you to avoid yet another substitution, where you could let:

$$u=2\theta\implies du=2d\theta=d(2\theta)$$

It simply means you would be integrating with respect to $2\theta$. :D

 

[karush]

Do you mean he identity of
$\sin^2\left({\theta}\right)=(1-\cos\left({2\theta}\right)) / 2$
I proceeded with this but the $\cos\left({2\theta}\right)$ got?
Sorry for delay on reply but it's the end of the quarter and I got overwhelmed

 

[MarkFL]

Yes, that's the identity I meant, so now we have:

$$I=\frac{1}{32}\int 1-\cos(4\theta)\,d(4\theta)=\frac{1}{32}\left(4\theta-\sin(4\theta)\right)+C$$

Now, back-substitute for $\theta$, perhaps using some trig. identities first. :D

 

[karush]

$\theta$?
Not sure how that was given?

 

[MarkFL]

$\theta$?
Not sure how that was given?

Take the substitution you used and solve it for $\theta$. :D

 

[karush]

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

This?

 

[MarkFL]

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

This?

Yes, that'd be it! :D

 

[karush]

$ \frac{1}{8}\left[\arcsin\left({x}\right)-x\right]+C$

I hope..

 

[MarkFL]

$ \frac{1}{8}\left[\arcsin\left({x}\right)-x\right]+C$

I hope..

No, you are going to need to use some double-angle identities...and a Pythagorean identity because your anti-derivative has $4\theta$ as the trig. argument...:D

 

[karush]

$\sin\left({4\theta}\right)=4\sin\left({\theta}\right)\cos^3\left({\theta}\right)
-4\sin^3\left({\theta}\right)\cos\left({\theta}\right)$

Do I proceed with this? that's expanding this a lot.

 

[MarkFL]

I would write:

$$\sin(4\theta)=2\sin(2\theta)\cos(2\theta)=4\sin(\theta)\cos(\theta)\left(1-2\sin^2(\theta)\right)$$

Now you just need to write $\cos(\theta)$ in terms of $\sin(\theta)$.

 

[karush]

I would write:

$$\sin(4\theta)=2\sin(2\theta)\cos(2\theta)=4\sin(\theta)\cos(\theta)\left(1-2\sin^2(\theta)\right)$$

Now you just need to write $\cos(\theta)$ in terms of $\sin(\theta)$.

this one? $\sin\left({\frac{\pi}{2}-\theta}\right)=\cos\left({\theta}\right)$