I'm having some trouble with my math homework again...:(

Use integration by parts to find the indefinite integral:

1) INT 1/(x*(lnx)^3) dx

2) INT x/sqroot(2+3x) dx

3) INT (x+sin3x)^2 dx

4) INT sin x^(1/3) dx

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- May 2nd 2007, 10:35 PMcupcakelova87Integration By Parts
I'm having some trouble with my math homework again...:(

Use integration by parts to find the indefinite integral:

1) INT 1**/**(x*(lnx)^3) dx

2) INT x**/**sqroot(2+3x) dx

3) INT (x+sin3x)^2 dx

4) INT sin x^(1/3) dx - May 2nd 2007, 10:38 PMecMathGeek
- May 2nd 2007, 10:43 PMecMathGeek
- May 2nd 2007, 10:51 PMcupcakelova87
Thanks ecmathgeek! Quick question...how did you get V for problem #2?

"Let dv = 1/sqrt(2 + 3x) dx <--> v = (2/3)sqrt(2 + 3x)" - May 2nd 2007, 10:54 PMecMathGeek
First, I would simplify the squared term:

INT x^2 + 2xsin3x + sin^2(3x) dx

Now this integration has three parts: x^2 (which is easy), 2xsin3x (which needs by-parts), sin^2(3x) (which needs trig identity).

INT x^2 dx + 2*INT xsin3x dx + INT sin^2(3x) dx

For INT x^2 dx:

INT x^2 dx

= 1/3*x^3

For INT 2xsin3x dx:

Let u = 2x <--> du = 2dx

Let dv = sin3x dx <--> v = -1/3*cos3x

=> -(2/3)xcos3x - 2/3*INT cos3x dx

= -(2/3)xcos3x - (2/9)sin3x

For INT sin^2(3x) dx:

Let sin^2(3x) = 1/2(1 - cos6x) <-- this is the half-angle formula for sin(3x) after squaring both sides.

1/2*INT 1 - cos6x dx

= 1/2*(x - 1/6*sin6x)

= (1/2)x - (1/12)sin6x

Recombining these integrations, we get:

(1/3)x^3 - (2/3)xcos3x - (2/9)sin3x + (1/2)x - (1/12)sin6x + C - May 2nd 2007, 10:57 PMecMathGeek
Good question. I skipped that step, but I'll explain it.

We have:

dv = 1/sqrt(2 + 3x) dx

To get v, we need to integrate both sides:

INT dv = INT 1/sqrt(2 + 3x) dx

The integration of dv is easy. The integration of 1/sqrt(2 + 3x) requires an additional substitution.

Let n = 2 + 3x <--> dn = 3 dx --> dx = 1/3 dn

1/3*INT 1/sqrt(n) dn

= 1/3*INT (n)^(-1/2) dn

= 1/3*(2/1)*n^(1/2)

= (2/3)sqrt(2 + 3x)

Therefore:

v = (2/3)sqrt(2 + 3x) - May 2nd 2007, 11:01 PMecMathGeek
Anytime I show something like:

Let dv = f(x) dx <--> v = F(x)

Where F(x) is the antiderivative of f(x), I recommend you try to figure out how I went from f(x) dx to F(x). Try doing the integration yourself. It should improve your ability to do integrations. - May 2nd 2007, 11:05 PMecMathGeek
Let u = sinx^(1/3) <--> du = 1/3*x^(-2/3)*sinx^(1/3) dx

Let dv = dx <--> v = x

=> xsinx^(1/3) - (1/3)*INT x^(-2/3)sinx^(1/3) dx

Let w = x^(1/3) <--> dw = (1/3)x^(-2/3) dx --> dx = 3*x^(2/3) dw

=> xsinx^(1/3) - INT sinw dw

= xsinx^(1/3) + cosw + C

= xsinx^(1/3) + cosx^(1/3) + C