Question

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...

Answer 1

A) `int \ e^(2x)/(1+e^x) \ dx = C + x/2+(3x^2)/8+(x^3)/6+(3x^4)/64 + ...`

B) `int \ xarcsinx \ dx = C + x^3/3+(x^5)/30 + ...`

Explanation:

We seek:
A) `I_1 = int \ e^(2x)/(1+e^x) \ dx`
B) `I_2 = int \ xarcsinx \ 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(x) = e^(2x)/(1+e^x) => f^((0))(0) = 1/2`

Differentiating wrt (steps omitted) `x`:

`f^((1))(x) = ((e^x+2)e^(2x))/(e^x+1)^2 => f^((1))(0) = 3/4`

Differentiating wrt again (steps omitted) `x`:

`f^((2))(x) = ((e^(2x)+3e^x+4)e^(2x))/(e^x+1)^3 => f^((2))(0) = 1`

Differentiating wrt again (steps omitted) `x`:

`f^((3))(x) = ((e^(3x)+4e^(2x)+5e^x+8)e^(2x))/(e^x+1)^3 => f^((3))(0) = 9/8`

Thus we get:

`f(x) = 1/2 + 3/4x + ((1)x^2)/(2) + (9/8x^3)/(6) + ...`

So we can write:

`I_1 = int \ {1/2+(3x)/4+(x^2)/2+(3x^3)/16 + ... } \ dx`
`\ \ \ = C + x/2+(3x^2)/8+(x^3)/6+(3x^4)/64 + ...`

Part (B)

with:

`f(x) = xarcsinx => f^((0))(0) = 0`

Differentiating wrt (steps omitted) `x`:

`f^((1))(x) = arcsinx+x/sqrt(1-x^2) => f^((1))(0) = 0`

Differentiating wrt again (steps omitted) `x`:

`f^((2))(x) = (2-x^2)/(1-x^2)^(3/2) => f^((2))(0) = 2`

Differentiating wrt again (steps omitted) `x`:

`f^((3))(x) = (x(4-x^2))/(1-x^2)^(5/2) => f^((3))(0) = 0`

Differentiating wrt again (steps omitted) `x`:

`f^((4))(x) = (4+13x^2-2x^4)/(1-x^2)^(7/2) => f^((4))(0) = 4`

Thus we get:

`f(x) = 0 + 0x + ((2)x^2)/(2) + (0x^3)/(6) + (4x^4)/(24) + ...`

So we can write:

`I_2 = int \ {x^2 + x^4/6 + ... } \ dx`
`\ \ \ = C + x^3/3+(x^5)/30 + ...`