How do you find the integral of #f(x)=2xsin4x# using integration by parts?

1 Answer
Shura
Feb 27, 2016

#int 2x sin(4x) dx=(-1)/2x cos(4x) + 1/8 sin(4x) + C#, where #C# is the constant of integration.

Explanation:

Let be #f(x) = 2x sin(4x)#.

We want to find #int 2x sin(4x) dx#. We know the integral of the function #sin# or #cos# so it would be good for us to get rid of #2x# by differentiating it.

Let's say #g(x) = 2x# and #h'(x) = sin(4x)#.

So #g'(x) = 2# and #h(x) = (-1)/4 cos(4x)#.

#int g(x) h'(x) dx = g(x)h(x) - int g'(x) h(x) dx# using integration by parts.

#int 2x sin(4x) dx = 2x*(-1)/4 cos(4x) - int 2 *(-1)/4 cos(4x) dx#

#=(-1)/2x cos(4x) + 1/2 int cos(4x) dx#

#=(-1)/2x cos(4x) + 1/2 * 1/4 sin(4x)#

#=(-1)/2x cos(4x) + 1/8 sin(4x)#

Therefore, #int 2x sin(4x) dx = (-1)/2x cos(4x) + 1/8 sin(4x) + C#, where #C# is the constant of integration.

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