# Confusing Trig/rational Integral

1. May 9, 2005

### Chaz706

$$\int (sin(t)-cos(t)) \sqrt{cos^2(t)-sin^2(t)} dt$$

Is there a trig idendity I can use? I've distributed that root to both terms to get:
$$\int sin(t) \sqrt{cos^2(t)-sin^2(t)} dt -$$ $$\int cos(t) \sqrt{cos^2(t)-sin^2(t)}$$

If I take one of the terms and integrate by parts, I'm trying to put $$u=\sqrt{cos^2(t)-sin^2(t)}$$ and $$dv= sin(t)$$ or $$dv= cos(t)$$ but that ugly root's derivative appears inside the $$\int vdu$$ part.

Is there a trig identity I'm missing, or some other tactic I could use? or could this just be really really ugly math?

2. May 9, 2005

### whozum

Its either

$$cos^2(t) - sin^2(t) = cos(2t)$$**

or

$$sin^2(t) - cos^2(t) = cos(2t)$$

3. May 9, 2005

### Chaz706

Whozum:
$$cos^2(t) - sin^2(t) = cos(2t)$$ is the correct identity (and thanks! It may just help!)

Your other identity: $$sin^2(t) - cos^2(t) = cos(2t)$$ is erroneous, but not by much. $$sin^2(t) - cos^2(t)$$ yields $$-cos(2t)$$

This could actually help as well, however. Thank you for posting both (even if one was inaccurate).

4. May 9, 2005

### dextercioby

You needn't that identity.

$$\int \left(\sin t-\cos t\right)\sqrt{\cos^{2}t-\sin^{2}t} \ dt = \int \sin t\sqrt{2\cos^{2}t-1} \ dt -\int \cos t\sqrt{1-2\sin^{2}t} \ dt$$

Can u take it from here (HINTbvious substitutions necessary) ?

Daniel.

5. May 9, 2005

### Chaz706

That would have been much easier dex, I think....

6. May 9, 2005

### whozum

Well it was one of the two, I justw anst sure which one. dex's way is easier though its basically the same thing.

7. May 10, 2005

### Chaz706

The Method is substitution, but then what's u? I'm guessing the basic trig function outside the root is du.

But when I derive the function within that root, I get something different. The derivative of either one ends up to be $$-4cos(t)sin(t)$$.

Am I choosing the wrong U? I'm integrating with respect to t, and simply inserting the extra parts onto by basic trig function isn't exactly kosher.

8. May 10, 2005

### dextercioby

Okay,let's take the first,i'll leave with the second,which is basically similar (maybe other substitution).

$$I=\int \sin t\sqrt{2\cos^{2}t-1} \ dt$$ (1)

becomes after the substitution $\cos t=u;\sin t \ dt= -du$ (2)

$$I=-\int \sqrt{2u^{2}-1} \ du$$(3)

Make the substitution

$$\sqrt{2} \ u=\cosh v;du=\frac{1}{\sqrt{2}} \sinh v \ dv$$(4)

Then

$$I=-\frac{\sqrt{2}}{2}\int \sinh^{2}v \ dv=-\frac{\sqrt{2}}{4}\int (\cosh 2v-1) \ dv=-\frac{\sqrt{2}}{8}\sinh 2v+\frac{\sqrt{2}}{4}v +C$$ (5)

And now invert the 2 substitutions

$$I=-\frac{\sqrt{2}}{8}\sinh \left(2 \ \mbox{arg}\cosh\left(\sqrt{2}\cos t\right)\right)+\frac{\sqrt{2}}{4}\mbox{arg}\cosh\left(\sqrt{2}\cos t\right) + C$$ (6)

I'm sure the other one will be simpler.

Daniel.

Last edited: May 10, 2005
9. May 10, 2005

### Chaz706

Didn't know that if you took $$cos(t) = u$$ that you could allow $$u^2$$ to simply be $$cos^2(t)$$.

Of course, I didn't think about that either.

10. May 10, 2005

### dextercioby

The other way around is incorrect,of course,but i simply squared.

Daniel.