# Using Integration by parts

1. Sep 1, 2004

### Whatupdoc

$$\int 4x cos(2x)$$

using integration by parts...

u=4x
du= 4
dv=cos(2x)
v=cos(x)sin(x)

using the formula uv - $$\int$$v*du....

4x*cos(x)sin(x) - $$\int$$cos(x)sin(x)*4x

hmm i cant seem to finish this problem, can someone help? and am i doing it correctly so far?

2. Sep 1, 2004

### Tide

Your expression for v is incorrect.

3. Sep 1, 2004

### arildno

DO NOT USE THIS HORRIBLE MIXTURE OF INTEGRATION BY PARTS AND INTEGRATION BY SUBSTITUTION!!
Use the following:
$$\int{uv'dx}=uv-\int{u'vdx}+C$$
Set:
$$u(x)=4x, v'(x)=\cos(2x)\to{v(x)}=\frac{1}{2}\sin(2x)$$

4. Sep 2, 2004

### Whatupdoc

ok so i plugged the parts into the formula
$$\int{uv'dx}=uv-\int{u'vdx}+C$$

$$4x*1/2*sin(2x) - \int 4*1/2*sin(2x)*dx + C$$

anti-deri. of sin is -cos that is why it's adding now...

4x*1/2*sin(2x) + 4x*1/4*cos(2x)
so... would that be the answer?

5. Sep 2, 2004

### Ethereal

Yes, it's the answer, try differentiating it.

EDIT: Error.

Last edited by a moderator: Sep 2, 2004
6. Sep 2, 2004

### Whatupdoc

the thing is, the answer is incorrect. i dont know what the correct answer is, but i know that isnt the answer. i submit my homework thru a online server and it checks your answer. what am i doing wrong?

7. Sep 2, 2004

### Ethereal

My deepest apologies. I must have been blind! The correct answer should be $$2xsin2x + cos2x$$ (written in Tex for clarity)

8. Sep 2, 2004

### Whatupdoc

thanks for the correct answer, but where did i go wrong?

9. Sep 2, 2004

### Ethereal

That x is not supposed to be there.

10. Sep 3, 2004

### Whatupdoc

$$4x*1/2*sin(2x) - \int 4*1/2*sin(2x)*dx + C$$

hmm i thought i was suppose to take the anti-derivative of 4 also.... can you tell me why i dont need to take the anti-derivative of the "4" after the intergral?

11. Sep 3, 2004

### Math Is Hard

Staff Emeritus
here are my thoughts..

$$\int 4*1/2*sin(2x)dx$$
is equivalent to
$$\int 2 *sin(2x)dx$$
is equivalent to
$$2* \int sin(2x)dx$$

you can always pull out the constant factor

Last edited: Sep 3, 2004
12. Sep 5, 2004

### HallsofIvy

What do you mean by "take the anti-derivative of 4 also"? Did you also think you should take the anti-derivative of the 1/2?
$\int{f(x)g(x)dx}$ is NOT $\int{f(x)dx}\int{g(x)dx}$.

In particular, if C is any constant (4, 1/2, whatever),
$$\int{Cf(x)dx}= C\int{f(x)dx}[/itex]. [tex]\int{4(1/2)sin(2x)dx}= \int{2sin(2x)dx}= 2\int{sin(2x)dx}= -cos(2x)+ C$$.

13. Sep 6, 2004

### Whatupdoc

that was very very very helpful, thank you sooo much. i always thought that $\int{f(x)g(x)dx}$ is equal to $\int{f(x)dx}\int{g(x)dx}$ .

ok one more thing, so what is $\int{f(x)g(x)dx}$ then? would i just take the anti-derv. of f(x)?

14. Sep 6, 2004

### Tide

No. Generally, you would have to find the antiderivative of $f(x)g(x)$. Knowing the antiderivative of one of them would permit you to use integration by parts.

15. Sep 6, 2004

### Ethereal

This is a little off-topic, but given a function, how do we know if it's anti-derivative exists?

16. Sep 6, 2004

### Muzza

I think all continous functions have anti-derivatives, because given a real continous function f defined on say [a, b], if we define A by

$$A(x) = \int^{x}_{a} f(t) dt$$

then A'(x) = f(x) (for all sensible values of x) (this is usually part of the proof of the fundamental theorem of (integral) calculus). So an anti-derivative exists (actually writing such a function in a closed form is a totally different matter though ;)).

Someone correct me if I'm wrong...

Last edited: Sep 6, 2004
17. Sep 6, 2004

### HallsofIvy

Every continuous function (in fact, every bounded function whose points of discontinuity form a set of measure 0) has an anti-derivative. Whether or not that anti-derivative is an "elementary" function depends on your definition of "elementary" function!

No, of course, not! You can't just ignore g! The way you integrate a product IS "integration by parts".

Remember the differentiation laws? IF (fg)'= f' g' THEN we could integrate both sides and get $\int{f(x)g(x)dx}=\int{f(x)g(x)dx}$.

But that's NOT true. What is true is the "product rule": (fg)'= f'g+ fg'.
That's the same as fg'= (fg)'- f'g. Integrating both sides,
$\intf(x)g'(x)dx= fg- \int g f' dx$.

Writing u= f, dv= g'dx, that is exactly $\int udv= uv- \int vdu$.

18. Sep 6, 2004

### Whatupdoc

i read your post many many times, and i kinda get what your saying, but i am also confused about some parts. im way better with examples, can you help me step by step and solve this problem:

$$\int 4x cos(2x)$$ <-- this problem is the one i asked for from my first post, i already got the answer, but i dont know how Ethereal got the answer.

let me show you how i tried to solve the problem.

using $$\int{uv'dx}=uv-\int{u'vdx}+C$$
u(x) = 4x
du(x) = 4dx
dv(x) = cos(2x)
v = 1/2sin(2x)

plugged in what i have...

$$4x*1/2sin(2x) \int 4*1/2sin(2x)$$

ok now i know that i need to solve $$\int 4*1/2sin(2x)$$
ok at this point, i am really stuck and confused. can someone help me step by step and explain to me what i really need to do? i know it's alot to ask for, but i really want to learn it.

Last edited: Sep 6, 2004
19. Sep 6, 2004

### Math Is Hard

Staff Emeritus
I think there's still some problems in your setup. The formula for int. by parts is:
$\int udv= uv- \int vdu$

Let u = 4x
dv = cos(2x)dx
du = 4 dx
v = 1/2 sin(2x)

$\int 4x cos(2x)dx = 4x (1/2 sin (2x)) - \int (1/2 sin (2x)) 4 dx$

which becomes:

$\int 4x cos(2x)dx = 2x sin (2x) - \int 2 sin (2x) dx$

which becomes

$\int 4x cos(2x)dx = 2x sin (2x) - 2\int sin (2x) dx$

the last integral is pretty easy to solve with a substitution for 2x. Does that help at all?

20. Sep 6, 2004

### Math Is Hard

Staff Emeritus
If you're stuck on the last part, show how you would integrate this:

$\int sin(2x)dx$

that should pinpoint where the problem is.