Question

How do you integrate `\int _ { 0} ^ { \frac { \pi } { 4} } 2x \sin x d x`?

Answer 1

The integral equals approximately `0.30`.

Explanation:

Use integration by parts.

Let `u = 2x` and `dv = sinx`. Then `du = 2dx` and `v = -cosx`

`I = int (u dv) = uv - int(v du)`

`I = int (2xsinx) = -2xcosx - int(-2cosx)dx`

`I = int (2xsinx) = -2xcosx + 2intcosx dx`

`I = int(2xsinx) = 2sinx - 2xcosx + C`

If we make this a definite integral, though, we have:

`I = [2sinx - 2xcosx]_0^(pi/4)`

`I = 2sin(pi/4) - 2(pi/4)cos(pi/4) - (2sin(0) - 2(0)cos(0))`

`I = 2/sqrt(2) - pi/(2sqrt(2))`

`I ~~ 0.30`

Hopefully this helps!