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September 16, 2014

September 16, 2014

Posted by **gerrard** on Saturday, May 21, 2011 at 1:43pm.

- calculus -
**MathMate**, Sunday, May 22, 2011 at 9:39pmAn interesting problem! In fact, it would make a good exam question, because you need to know four rules to calculate the integral, or equivalent to four short questions.

(cos(4x)+2x^2)(sin(4x)-x)

=cos(4x)sin(4x)-xcos(4x)+2x²sin(4x)-2x³

1. cos(4x)sin(4x)

use substitution

u=sin(4x)

du=4cos(4x)dx

so

∫cos(4x)sin(4x)dx

=(1/4)∫udu (use the power rule)

2. -xcos(4x)

Use integration by parts:

∫-xcos(4x)dx

=-(1/4)xsin(4x)+(1/4)∫sin(4x)

3. 2x²sin(4x)

Use integration by parts, similar to (2) above.

4. -2x³

use the power rule.

Post your answer for a check if you wish. Do not forget the constant of integration (C).

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