Hyperbolic Headaches..

**minusb** · November 20th 2010, 01:47 PM

Hi guys I'm trying to work through some problems getting ready for an exam and I came across this problem that doesn't make sense to me...

Consider $\displaystyle \int x(1+x^2)dx$

If I integrate this I get $1/4(x^4) + 1/2(x^2)$

It says if we substitute x = sinh(u) we get $\displaystyle \int sinh(u)cosh^3(u)du$

Where on earth does that cosh^3 come from? I know that $1+sinh^2(u)$ gives us $cosh^3(u)du.$

Also if I integrate this I get $\displaystyle \int sinh(u)cosh^3(u)du$ I get $x^4/4 + 2x^2/4 + 1/4$ . Why are they not equal?

Thoughts appreciated?

**dwsmith** · November 20th 2010, 01:56 PM

$\displaystyle dx=cosh(u)du$

Therefore, $\dispalystyle sinh(u)*(1+sinh(u)^2)cosh(u)=sinh(u)*cosh(u)^2*cos h(u)=sinh(u)cosh(u)^3$

Integrating yields:

$\displaystyle\frac{cosh(u)^4}{4}$

$\displaystyle u=sinh^{-1}(x)\rightarrow\frac{cosh^4(sinh^{-1}(x))}{4}$

**dwsmith** · November 20th 2010, 02:04 PM

When integrating, we have a constant of integration with indefinite integrals. $\frac{1}{4}$ is a constant but we would still have a plus C. $C+\frac{1}{4}=C_2$ .

Integral without substitution: $\displaystyle\frac{x^2}{2}+\frac{x^4}{4}+C$

With: $\displaystyle\frac{x^2}{2}+\frac{x^4}{4}+\frac{1}{ 4}+C=\frac{x^2}{2}+\frac{x^4}{4}+C_2$

**TheCoffeeMachine** · November 20th 2010, 02:12 PM

What a title!

Originally Posted by minusb

Also if I integrate this I get $\displaystyle \int sinh(u)cosh^3(u)du$ I get $x^4/4 + 2x^2/4 + 1/4$ . Why are they not equal?

I've not checked your calculations, but they don't have to be. When you integrate some function f(x) over
an indefinite interval, infinite number of functions that satisfy what you get (but there might be exceptions
to this) -- what you find is the set of all anti-derivatives. That's why we have the constant of integration.

You have:

$\displaystyle I_{1} = \frac{1}{4}x^4+\frac{1}{2}x^2+k_{1}$

$\displaystyle I_{2} = \frac{1}{4}x^4+\frac{1}{2}x^2+\frac{1}{4}+k_{2}$

$k_{1}$ and $k_{2}$ are constants, of course. But so is $\frac{1}{4}+k_{2}$ and $\frac{1}{4}+k_{2}+k_{1}$ , call this $K$ .

So $\displaystyle I = \frac{1}{4}x^4+\frac{1}{2}x^2+K$

An indefinite integral is really a big function that accounts for all the functions whose derivative is the integrated function!

**minusb** · November 20th 2010, 02:19 PM

That makes sense. Thank you fellas. Keep up the great work.

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