# How do you integrate int e^x tan sqrtx dx  using integration by parts?

May 14, 2016

You have to apply the definition: $\int u \mathrm{dv} = u v - \int v \mathrm{du}$. Nonetheless, this integral does seem solvable analytically.

#### Explanation:

If you apply it once:

Choose: $\mathrm{dv} = \exp \left(x\right) \mathrm{dx}$, and $u = \tan \sqrt{x}$
Which gives you: $v = \exp \left(x\right)$, and $\mathrm{du} = {\sec}^{2} \frac{\sqrt{x}}{2 \sqrt{x}} \mathrm{dx}$

Thus:

$\int \exp \left(x\right) \tan \sqrt{x} \mathrm{dx} = \exp \left(x\right) \cdot {\sec}^{2} \frac{\sqrt{x}}{2 \sqrt{x}} - \int \exp \left(x\right) {\sec}^{2} \frac{\sqrt{x}}{2 \sqrt{x}} \mathrm{dx}$

Unfortunately, the more you go, the more complicate it gets, neither if you make $u = \exp \left(x\right)$, and $\mathrm{dv} = \tan \sqrt{x} \mathrm{dx}$, the exponential "locks" the integrals, making them cumbersome. Maybe this integral is no solvable by integration by part, I am not sure even analytically.

A possible way to solve it is by Taylor expansion, expand $\tan \sqrt{x}$, which is the problematic part of the integral. I will try to add a second answer using this method.