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September 30, 2014

September 30, 2014

Posted by **Anonymous** on Wednesday, November 7, 2012 at 10:15am.

e^2x Cos(4x) dx

Question 2)

x sin(x) dx

Question 3)

x^2 sin(x) dx

- Maths - Integration by Parts -
**Steve**, Wednesday, November 7, 2012 at 10:31amI'll do #1, you can show us where you get stuck on the others.

Let I = ∫e^2x cos4x dx

u=cos 4x

du = -4sin4x dx

dv = e^2x dx

v = 1/2 e^2x

I = uv - ∫v du

= 1/2 e^2x cos4x - ∫-2e^2x sin4x dx

Now do it again, letting u = sin4x

I = 1/2 e^2x cos4x + (e^2x sin4x - ∫4e^2x cos4x dx)

= 1/2 e^2x cos4x + e^2x sin4x - 4I

5I = 1/2 e^2x cos4x + e^2x sin4x

I = 1/10 e^2x (cos4x + 2sin4x)

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