Confusing Trig/rational Integral

[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?

 

[whozum]

Its either

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

or

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

 

[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).

 

[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.

 

[Chaz706]

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

 

[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.

 

[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.

 

[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.

 

[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.

 

[dextercioby]

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

Daniel.