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Chapter 7: Techniques of Integration

Ok! Chapter 7! Here we go!

Now I'm starting on the infamous "Integration by parts," and when the book says "every differentiation rule has a corresponding integration rule," this really seems to clear a lot of things up for me, and makes it easier for me to really get a grasp on integration techniques. I think, "of course! Why didn't I think of that before?"

So the formula for integration by parts is

∫ f(x) g '(x) dx = f(x) g(x) - ∫ g(x) f '(x) dx

or

∫ u dv = uv - ∫ v du

Its so simple.

Now, I've done a few problems, but I really get the feeling that a couple of these sin and cos problems I did are wrong. Here they are:

I'm sure this one is right:

Evaluate the integral:

6) ∫ t sin 2t dt = A

My work:

u = t

dv = sin 2t

du = dt

v = (-1/2)cos 2t

A = (-t/2) cos 2t + ∫ (1/2) cos 2t

A = (-t/2) cos 2t + (1/4) sin 2t

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7) ∫ x<sup>2</sup>sin πx dx

My work:

∫ x<sup>2</sup>sin πx dx = A

u = x<sup>2</sup>

dv = sin πx

du = 2x

v = (-1/π) cos πx

A = (-x<sup>2</sup>/π ) cos πx - ∫ (-2x/π) cos πx dx

∫ (-2x/π) cos πx dx= B

w = -2x/π

dz = cos πx

dw = -2/π

z = (1/π)sin πx

B = (-2x/π<sup>2</sup>) sin πx + ∫ (2/π<sup>3</sup>) sin π x dx

B = (-2x/π<sup>2</sup>) sin πx - (2/π<sup>3</sup>) cos πx

A = (-x<sup>2</sup>/π ) cos πx - B

so

∫ x<sup>2</sup>sin πx dx = (-x<sup>2</sup>/π ) cos πx - ( (-2x/π<sup>2</sup>) sin πx - (2/π<sup>3</sup>) cos πx )

or

∫ x<sup>2</sup>sin πx dx = (-x<sup>2</sup>/π ) cos πx + (2x/π<sup>2</sup>) sin πx+ (2/π<sup>3</sup>) cos πx

The book answer is:

(-x<sup>2</sup>/π) cos πx + (2x/π<sup>2</sup>) sin πx + (2/π<sup>3</sup>) cos πx

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8 ) ∫ x<sup>2</sup> cos mx dx

My work:

∫ x<sup>2</sup> cos mx dx = A

u = x<sup>2</sup>

dv = cos mx

du = 2x

v = (1/m) sin mx

A = x<sup>2</sup> sin mx - ∫ (2x/m) sin mx dx

∫ (2x/m) sin mx dx = B

w = (2x/m)

dz = sin mx

du = 2/m

z = (-1/m) cos mx

B = (-2x/m<sup>2</sup> ) cos mx - ∫ (-2/m<sup>2</sup> ) cos mx

B = (-2x/m<sup>2</sup>) cos mx + (2/m<sup>3</sup>) sin mx

A = x<sup>2</sup> sin mx - ( (-2x/m<sup>2</sup>) cos mx + (2/m<sup>3</sup>) sin mx )

A = x<sup>2</sup> sin mx + (2x/m<sup>2</sup>) cos mx - (2/m<sup>3</sup>) sin mx

It seems to me that a general technique to use with integration by parts is to choose your "u" as a part of the integral that has a positive exponent and your "dv" as a part that has a negative exponent.