# Thread: Antiderivative involving hyp. functions.

1. ## Antiderivative involving hyp. functions.

I haven't much experience with hyperbolic functions, having never learned them, but the result of this integral seems to involve one:

$\int{\sqrt{1+u^2} du}$

Their solution is:

$\frac{1}{2}u\sqrt{1+u^2}+\frac{1}{2}\text{arsinh}{ u}$

(What's the proper latex command for arsinh?)

I'm not sure how they derived that.

2. They've used the identity $cosh^2x-sinh^2x=1$.

Remember that $\frac{d}{dx}cosh(x)=sinh(x)$ and $\frac{d}{dx}sinh(x)=cosh(x)$.

3. Ok, so using that identity, would they then have done:

$\int{\sqrt{u^2+\cosh^2(u)-\sinh^2(u)}}du$ ?

If so, would this be integrated by parts?

But I'm still not sure how they ended up with an Inverse Hyperbolic Sine (arsinh). Since the $\int{\sinh(x)}dx = \cosh(x)$, right? I don't see where arsinh comes into this.

4. Solve $
\int{\sqrt{1+u^2} du}
$
.
Let $u=sinh(x)$. Therefore $\frac{du}{dx}=cosh(x)$.

The integral becomes: $\int\sqrt{1+sinh^2(x)} cosh(x) \ dx$

$\int \sqrt{cosh^2(x)} cosh(x) \ dx$

$\int cosh^2(x) \ dx$

Now use the identity $cosh^2(x)+sinh^2(x)=cosh(2x)$ and $cosh^2(x)-sinh^2(x)=1$ to get $2cosh^2(x)-1=cosh(2x)$.

Since we require $cosh^2(x)$ we can rearrange this to get $cosh^2(x)=\frac{cosh(2x)+1}{2}$.

We now need to integrate $\frac{1}{2}\int cosh(2x)+1dx=\frac{1}{2}(\frac{1}{2}sinh(2x)+x+C)$

$sinh(2x)=2sinh(x)cosh(x)$ but using $cosh^2(x)-sinh^2(x)=1$ we get that $cosh(x)=\sqrt{1+sinh^2(x)}$. Combining this with the other identity gives $sinh(2x)=2sinh(x)\sqrt{1+sinh^2(x)}$.

We also know that $x=arsinh(u)$ from how we defined u at the very start.

Therefore $\frac{1}{2}(\frac{1}{2}sinh(2x)+x)+K$ $=\frac{1}{2}(\frac{1}{2}2sinh(x)\sqrt{1+sinh^2(x)} +x)+K=\frac{1}{2}u\sqrt{1+u^2}+\frac{1}{2}arsinh(u )+K$.

In this case, they've taken $K=C=0$.

5. Thank you very much for the explanation