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

Question

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

Scarlett Allison · Accepted Answer

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 + ... # 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 + ... #

Christopher Allers · Answer

undefined To solve $ \int \frac{e^{2x}}{1+e^{4x}}dx $ using infinite series, we can expand $ \frac{1}{1+e^{4x}} $ as a geometric series and then integrate each term. Similarly, for $ \int x \arcsin(x)dx $, we'll express $ \arcsin(x) $ as its Maclaurin series and integrate the resulting series term by term.

For $ \int \frac{e^{2x}}{1+e^{4x}}dx $:

1. Expand $ \frac{1}{1+e^{4x}} $ as a geometric series:
   $$ \frac{1}{1+e^{4x}} = \frac{1}{1+e^{2x}\cdot e^{2x}} = \frac{1}{1+e^{2x}}\cdot\frac{1}{1+e^{2x}} $$
   Expand $ \frac{1}{1+e^{2x}} $ using the geometric series formula:
   $$ \frac{1}{1+e^{2x}} = \sum_{n=0}^{\infty} (-1)^n e^{2nx} $$

2. Integrate each term of the series:
   $$ \int e^{2nx} dx = \frac{1}{2n} e^{2nx} + C $$

3. Plug back the expression of the series and integrate term by term:
   $$ \int \frac{e^{2x}}{1+e^{4x}}dx = \int \sum_{n=0}^{\infty} (-1)^n e^{2nx} dx $$
   $$ = \sum_{n=0}^{\infty} (-1)^n \int e^{2nx} dx $$

For $ \int x \arcsin(x)dx $:

1. Express $ \arcsin(x) $ as its Maclaurin series:
   $$ \arcsin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)2^n} x^{2n+1} $$

2. Integrate each term of the series:
   $$ \int x^{2n+1} dx = \frac{1}{2n+2} x^{2n+2} + C $$

3. Plug back the expression of the series and integrate term by term:
   $$ \int x \arcsin(x)dx = \int x \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)2^n} x^{2n+1} dx $$
   $$ = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)2^n} \int x^{2n+1} dx $$

Please note that both of these series representations may converge only for certain values of $ x $.