What is #int xsinx#?
1 Answer
Nov 8, 2015
#int x sin x dx = sin(x) - x cos(x) + C#
Explanation:
Use the following:
#d/dx x = 1#
#d/dx sin(x) = cos(x)#
#d/dx cos(x) = -sin(x)#
#d/dx (f(x)g(x)) = f'(x)g(x) + f(x)g'(x)#
To find:
#d/dx (sin(x) - x cos(x) + C) = color(red)(cancel(color(black)(cos(x)))) - color(red)(cancel(color(black)(cos(x)))) + x sin(x) = x sin(x)#
Hence:
#int x sin x dx = sin(x) - x cos(x) + C#
