computing integrate of exponential divide by generalization trigonometry

Question

The integrate:

$$ \int{\frac{e^x}{\sqrt{4-e^{2x}}}dx} $$

My steps: $u=e^x$ and $du=e^xdx$ get this integrate:

$$ \int{\frac{du}{\sqrt{4-ue^2}}} $$

But for to apply the propertie $\int{\frac{du}{\sqrt{1+u²}}}= arc sin(\frac{u}{a})$ I don't know who is and $u$. (I know that $a$ is 2)

Answer 1

Let $u = \dfrac{e^x}2\implies2\,\mathrm du = e^x\mathrm dx$. Therefore,

$$\int\dfrac{e^x}{\sqrt{4 - e^{2x}}}\,\mathrm dx\equiv\int\dfrac2{\sqrt{4 - 4u^2}}\,\mathrm du = \int\dfrac1{\sqrt{1 - u^2}}\,\mathrm du$$

This is a standard integral for $\arcsin(u)$. That is,

$$\int\dfrac1{\sqrt{1 - u^2}}\,\mathrm du = \arcsin(u) + \mathrm{constant}$$

Undo substitution.

$$\int\dfrac{e^x}{\sqrt{4 - e^{2x}}}\,\mathrm dx = \arcsin\left(\dfrac{e^x}2\right) + \mathrm{constant}$$

Answer 2

You can also use that for $a>0$ $$\int\frac{1}{\sqrt{a^2-t^2}}\,dt=\arcsin\left(\frac{t}{a}\right)+C $$ Then $$\int\frac{1}{\sqrt{4-u^2}}\,du=\arcsin\left(\frac u2\right)+C $$ For a proof of that fact, simply differentiate $\arcsin\left(\frac{t}{a}\right)$: $$\left(\arcsin\left(\frac{t}{a}\right)\right)'=\frac 1a\frac{1}{\sqrt{1-\frac{t^2}{a^2}}}=\frac{1}{\sqrt{a^2-t^2}} $$