# Thread: Integration By Parts

1. ## Integration 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

2. Originally Posted by cupcakelova87
Use integration by parts to find the indefinite integral:
1) INT 1/(x*(lnx)^3) dx
Integration by parts for this? Substitution would work better here.

Let u = lnx <--> du = 1/x dx --> dx = x du

INT 1/(x*u^3)*x du
= INT 1/u^3 du
= -1/(2u^2) + C
= -1/(2(lnx)^2) + C

3. Originally Posted by cupcakelova87
Use integration by parts to find the indefinite integral:

2) INT x/sqroot(2+3x) dx
Let u = x <--> du = dx
Let dv = 1/sqrt(2 + 3x) dx <--> v = (2/3)sqrt(2 + 3x)

=> (2/3)x*sqrt(2 + 3x) - INT (2/3)sqrt(2 + 3x) dx
= (2/3)x*sqrt(2 + 3x) - (4/27)sqrt(2 + 3x)^3 + C

4. 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)"

5. Originally Posted by cupcakelova87
Use integration by parts to find the indefinite integral:

3) INT (x+sin3x)^2 dx
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

6. Originally Posted by cupcakelova87
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)"
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)

7. 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.

8. Originally Posted by cupcakelova87
Use integration by parts to find the indefinite integral:

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