I'm supposed to prove the following by using the hyperbolic sine double angle identity: $\sinh(2x)=2\sinh x\cosh x$ and archsinhx formula: $\ln(x+\sqrt{x^2+1})$ but can't seem to figure out the steps.
Prove: $\int\sqrt{x^2+a^2}\,dx = \frac{a^2}2 \ln\left(x+\sqrt{x^2+a^2}\right) + \frac{x}2 \sqrt{x^2+a^2} + C$
So far what I have is:
$$\int\sqrt{x^2+a^2}dx = a^2\int\cosh^2udu\ \ (\text{substitute }x=\operatorname{asinh}u)$$
$$ = \int(1+\sinh^2u)\ du = \int(\cosh(2u)-\cosh^2u)\ du = a^2\left(\frac{u}2-\frac14\sinh(2u)\right)$$
After that I'm just stuck. I've tried substituting $x$ back but no matter how I go about it, I can't seem to drive the proof. Can someone give me a nudge?
Best Answer
Use the double angle identity for $\cosh$. We have $$\cosh 2u=2\cosh^2 u-1,$$ and therefore $$\cosh^2 u=\frac{1}{2}(1+\cosh 2u)$$ Integrate. We get $$\frac{u}{2}+\frac{1}{4}\sinh 2u+C=\frac{u}{2}+\frac{1}{2}\cosh u\sinh u +C.$$ Now the back substitution should go nicely.