# Hyperbolic Substituion. I am wrong?

1. Aug 7, 2011

### flyingpig

1. The problem statement, all variables and given/known data

$$\int \;\sinh(2x) \cosh(2x) dx$$

3. The attempt at a solution

I let u = sinh(2x), du = 2cosh(2x)dx

Integrating I shuold get

$$\frac{1}{4} sinh^2 (2x) + C$$

But http://www.wolframalpha.com/input/?i=Integrate[cosh%282x%29%28sinh%282x%29%29%2Cx]

says I need to let u = 2x first. Why? I don't think i am wrong, my algebra looked correct.

2. Aug 7, 2011

### HallsofIvy

Staff Emeritus
No, you don't need to do that, you just can.

If you let u= 2x, then, of course, dx= (1/2) du so your integral becomes
$$\frac{1}{2}\int sinh(u)cosh(u)du$$
Then let v= sinh(u) so that dv= cosh(u)du and the integral becomes
$$\frac{1}{2}\int v dv= \frac{1}{4}v^2+ C= \frac{1}{4}sinh^2(u)+ C= \frac{1}{4}sinh^2(2x)+ C$$
just as you have.
That's just using two consecutive substitutions rather than combining them as you did.

Of course, you could do it the other way two: let u= cosh(2x) so that du= 2sinh(x)dx and the integral becomes
$$\frac{1}{2}u^2du= \frac{1}{4}cosh^2(2x)+ D$$

Because $cosh^2(2x)= sinh^2(2x)+ 1$ that is exactly the same thing with a difference constant.

I believe your wolframalpha reference gives the integral as
$$\frac{1}{8}cosh(4x)+ Constant$$
which looks very different but is actually the same as your result.

Remember that
$$sinh(x)= \frac{e^x- e^{-x}}{2}$$

Squaring that:
$$sinh^2(x)= \frac{(e^x)^2- 2e^xe^{-x}+ (e^{-x})^2}{4}= \frac{e^{2x}- 2+ e^{-2x}}{4}$$
$$= \frac{1}{2}\frac{e^{2x}+ e^{-2x}}{2}- \frac{1}{2}= \frac{1}{2}cosh(2x)- \frac{1}{2}$$
so that
$$\frac{1}{4}sinh^2(2x)= \frac{1}{8}cosh(4x)- \frac{1}{8}$$
differing from the original form only by a constant.

Last edited: Aug 7, 2011
3. Aug 7, 2011

### flyingpig

Oh right, we are doing an indefinite integral here.