Link to book index:
http://www.thescienceforum.com/Calcu...ions-8994t.php
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
--------------------------------------------------------------
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.