# Solve \int(e^(2x))/(1+e^(4x))dx, OR \intx\arcsin(x)dx using infinite series?

## I have the solution using normal integration, but considering the rest of the worksheet problems revolve around using infinite series to solve integrals...

May 14, 2018

A) $\int \setminus {e}^{2 x} / \left(1 + {e}^{x}\right) \setminus \mathrm{dx} = C + \frac{x}{2} + \frac{3 {x}^{2}}{8} + \frac{{x}^{3}}{6} + \frac{3 {x}^{4}}{64} + \ldots$

B) $\int \setminus x \arcsin x \setminus \mathrm{dx} = C + {x}^{3} / 3 + \frac{{x}^{5}}{30} + \ldots$

#### Explanation:

We seek:
A) ${I}_{1} = \int \setminus {e}^{2 x} / \left(1 + {e}^{x}\right) \setminus \mathrm{dx}$
B) ${I}_{2} = \int \setminus x \arcsin x \setminus \mathrm{dx}$

Using infinite Series.

Part (A)

We can form a power series for the integrand using the Maclaurin Series, so we have:

 f(x) =f^((0))(0) + f^((1))(0)x + (f^((2))(0)x^2)/(2!) + (f^((3))(0)x^3)/(3!) + ...

So, with:

$f \left(x\right) = {e}^{2 x} / \left(1 + {e}^{x}\right) \implies {f}^{\left(0\right)} \left(0\right) = \frac{1}{2}$

Differentiating wrt (steps omitted) $x$:

${f}^{\left(1\right)} \left(x\right) = \frac{\left({e}^{x} + 2\right) {e}^{2 x}}{{e}^{x} + 1} ^ 2 \implies {f}^{\left(1\right)} \left(0\right) = \frac{3}{4}$

Differentiating wrt again (steps omitted) $x$:

${f}^{\left(2\right)} \left(x\right) = \frac{\left({e}^{2 x} + 3 {e}^{x} + 4\right) {e}^{2 x}}{{e}^{x} + 1} ^ 3 \implies {f}^{\left(2\right)} \left(0\right) = 1$

Differentiating wrt again (steps omitted) $x$:

${f}^{\left(3\right)} \left(x\right) = \frac{\left({e}^{3 x} + 4 {e}^{2 x} + 5 {e}^{x} + 8\right) {e}^{2 x}}{{e}^{x} + 1} ^ 3 \implies {f}^{\left(3\right)} \left(0\right) = \frac{9}{8}$

Thus we get:

$f \left(x\right) = \frac{1}{2} + \frac{3}{4} x + \frac{\left(1\right) {x}^{2}}{2} + \frac{\frac{9}{8} {x}^{3}}{6} + \ldots$

So we can write:

${I}_{1} = \int \setminus \left\{\frac{1}{2} + \frac{3 x}{4} + \frac{{x}^{2}}{2} + \frac{3 {x}^{3}}{16} + \ldots\right\} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = C + \frac{x}{2} + \frac{3 {x}^{2}}{8} + \frac{{x}^{3}}{6} + \frac{3 {x}^{4}}{64} + \ldots$

Part (B)

with:

$f \left(x\right) = x \arcsin x \implies {f}^{\left(0\right)} \left(0\right) = 0$

Differentiating wrt (steps omitted) $x$:

${f}^{\left(1\right)} \left(x\right) = \arcsin x + \frac{x}{\sqrt{1 - {x}^{2}}} \implies {f}^{\left(1\right)} \left(0\right) = 0$

Differentiating wrt again (steps omitted) $x$:

${f}^{\left(2\right)} \left(x\right) = \frac{2 - {x}^{2}}{1 - {x}^{2}} ^ \left(\frac{3}{2}\right) \implies {f}^{\left(2\right)} \left(0\right) = 2$

Differentiating wrt again (steps omitted) $x$:

${f}^{\left(3\right)} \left(x\right) = \frac{x \left(4 - {x}^{2}\right)}{1 - {x}^{2}} ^ \left(\frac{5}{2}\right) \implies {f}^{\left(3\right)} \left(0\right) = 0$

Differentiating wrt again (steps omitted) $x$:

${f}^{\left(4\right)} \left(x\right) = \frac{4 + 13 {x}^{2} - 2 {x}^{4}}{1 - {x}^{2}} ^ \left(\frac{7}{2}\right) \implies {f}^{\left(4\right)} \left(0\right) = 4$

Thus we get:

$f \left(x\right) = 0 + 0 x + \frac{\left(2\right) {x}^{2}}{2} + \frac{0 {x}^{3}}{6} + \frac{4 {x}^{4}}{24} + \ldots$

So we can write:

${I}_{2} = \int \setminus \left\{{x}^{2} + {x}^{4} / 6 + \ldots\right\} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = C + {x}^{3} / 3 + \frac{{x}^{5}}{30} + \ldots$