What is f(x) = int (2x-xe^x)(-xe^x+secx) dx if f(0 ) = 5 ?

1 Answer
Apr 23, 2018

Can't be perfectly determined.

Explanation:

First, Integrate.

So, int (2x - xe^x)(-xe^x + secx)dx

= int (-2x^2e^x -x^2e^(2x) + 2xsec x - xe^xsecx)dx

= -2int x^2e^xdx -int x^2e^(2x)dx +2 int xsecx dx -int xe^xsecxdx

Now We have to Integrate By Parts.

Assume I_1 = -2 int x^2e^xdx,

I_2 = -int x^2e^(2x)dx

I_3 = 2int x sec xdx

I_4 = -intxe^xsecxdx

Now The Problem is,

I_1 and I_2 can be found, but I_3 and I_4 can't be expressed by elementary functions. These integrals are undeterminable.

So, Get the I_1 first.

I_1 = -2[x^2inte^xdx - int {d/dx(x^2)inte^x dx}dx]

= -2 [x^2e^x - 2 intxe^xdx]

= -2[x^2e^x - 2{x inte^xdx - int (d/dx(x) int e^xdx)dx}]

= -2[x^2e^x - 2{xe^x - inte^xdx}]

= -2[x^2e^x - 2{xe^x -e^x}]

= -2[x^2e^x - 2xe^x +2e^x]

= -2e^x[x^2 - 2x + 2]

Now, I_2.

I_2 = -2[x^2inte^(2x)dx - int {d/dx(x^2)inte^(2x) dx}dx]

= -2 [1/2x^2e^(2x) - 2 intxe^(2x)dx]

= -2[1/2x^2e^(2x) - 2{x inte^(2x)dx - int (d/dx(x) int e^(2x)dx)dx}]

= -2[1/2x^2e^(2x) - 2{1/2xe^(2x) - inte^(2x)dx}]

= -2[1/2x^2e^(2x) - 2{1/2xe^(2x) -1/2e^(2x)}]

= -2[1/2x^2e^(2x) - xe^(2x) +e^(2x)]

= -2e^(2x)[1/2x^2 - x + 1]

So, The Entire Integral is :

int(2x - xe^x)(-xe^x + sec x) dx

= -2e^x[x^2 - 2x + 2] -2e^(2x)[1/2x^2 - x + 1] + 2 intxsecxdx - intxe^xsecxdx

Hope this helps, but It won't, most probably.