Wednesday

July 23, 2014

July 23, 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).

**Related Questions**

Integral - That's the same as the integral of sin^2 x dx. Use integration by ...

trig integration - s- integral endpoints are 0 and pi/2 i need to find the ...

Calc BC - 1. Find the indefinite integral. Indefinite integral tan^3(pix/7)sec^2...

Calculus - Hello Everyone, I need help with Calc II. 1. Integral from 0 to 1 of...

calc - find integral using table of integrals ) integral sin^4xdx this the ...

Calculus II - Evaluate the integral using method of integration by parts: (...

Calculus - I need help with this integral. w= the integral from 0 to 5 24e^-6t ...

Calculus - "Evaluate the following indefinite integral using integration by ...

Calculus - Hello, I have some calculus homework that I can't seem to get started...

Calc - evaluate the indefinite integral: x^13(cos(x^7))dx I've tried using ...