1. Integrate dx/root(4+root(x))
Let's try it this way. Hope this works.
\(\displaystyle \int(\frac{1}{\sqrt{4+sqrt{x}}})dx\)
\(\displaystyle u^{2}=4+x^{\frac{1}{2}}\)
\(\displaystyle 2udu=\frac{1}{2sqrt{x}}dx, u^{2}-4=sqrt{x}\)
\(\displaystyle 4udu=\frac{1}{sqrt{x}}dx\)
\(\displaystyle (4u^{3}-16u)du=dx\)
\(\displaystyle \frac{(4u^{3}-16u)}{u}=(4u^{2}-16)\)
Now Integrate:
\(\displaystyle \int(4u^{2}-16)du\)
=\(\displaystyle \frac{4}{3}u^{3}-16u\)
=\(\displaystyle (\frac{4}{3})(4+\sqrt{x})^{\frac{3}{2}}-16\sqrt{4+\sqrt{x}}\)
Whatcha think?. Although it may not appear to, I believe this answer is equivalent to what Stapel posted.
May as well finish simplifying into the other form.
\(\displaystyle \frac{4}{3}((4+\sqrt{x})^{\frac{3}{2}})-16\sqrt(4+sqrt{x})\)
\(\displaystyle \frac{4}{3}(4+\sqrt{x})(\sqrt{4+\sqrt{x}})-16\sqrt{4+\sqrt{x}}\)
\(\displaystyle (\frac{16}{3}+\frac{4}{3}\sqrt{x})\sqrt{4+\sqrt{x}}-16\sqrt{4+\sqrt{x}}\)
\(\displaystyle \frac{16}{3}\sqrt{4+\sqrt{x}}+\frac{4}{3}\sqrt{x}\sqrt{4+\sqrt{x}}-16\sqrt{4+\sqrt{x}}\)
\(\displaystyle \frac{4}{3}\sqrt{x}\sqrt{4+\sqrt{x}}-\frac{32}{3}\sqrt{4+\sqrt{x}}\)
\(\displaystyle \frac{4}{3}(\sqrt{x}\sqrt{4+\sqrt{x}}-8\sqrt{4+\sqrt{x}})\)
\(\displaystyle \frac{4}{3}(\sqrt{x}-8)(\sqrt{4+\sqrt{x})\)
2. Verify:
(coshx-sinhx)^7=cosh(7x)-sinh(7x) (hyperbolic functions)
Are you allowed to use the hyperbolic identities:
\(\displaystyle \frac{e^{x}+e^{-x}}{2}=cosh(x)\) and \(\displaystyle \frac{e^{x}-e^{-x}}{2}=sinh(x)\)
.
On problem one, I thought about rationalizing the denominator and then using the integration of arctan, but that would get a bit messy.
On problem two, I though about a binomial expansion, but that would get quite messy as well.
Any ideas?
