can anyone help me with calculus integration?

Question

i have a lot of vacation homework and when i got home i realized i don't remember how to do intregration by parts can some one explain it to me.

Answer 1

I can try, sure.

Integration by parts is mainly used for finding integrals of products. For example, the integral of xsinx or the integral of
x[ln(x)]. How it works is that you work with two variables, u and v, and you let u = [one part of the product] and dv = [other part of the product. You then solve for du by taking the derivative of u, and you solve for v by taking the integral of dv.

I'll show you an example.

Integral (x sin(x) dx)

Let u = x. dv = sin(x) dx
du = dx. v = -cos(x)

[Note; when taking the "integral" it's not the general integral so you don't need to add C]

Having u and v, du and dv, Integration by parts relies on the formula below:

uv - Integral (v du).

Plugging our values in, we'd get

-xcos(x) - Integral (-cos(x) dx)

Note that the second part is now integrable. Let's simplify this a bit, shall we? First things first, we'll put out the - in front of the cos(x), bring it out of the integral, and then make it a +.

-xcos(x) + Integral (cos(x) dx)

Taking the integral, we now have

-xcos(x) + sin(x) + C

This is just one example; using this to help you along, I'd highly suggest you read more examples in your textbook, as explaining everything online isn't easy.

Something else to note is to be careful what you choose for your u and for your dv. What you want to choose for u is what gets simpler. Suppose we had done the following:

For Integral (x sin(x) dx)

Let u = sin(x). dv = x dx
du = cos(x). v = (x^2)/2

uv - Integral (v du)

[sin(x)] [(x^2)/2] - Integral ( [x^2]/2 [cos(x)]) dx

And, obviously, the integral is now more difficult to solve than the original equation. That's why you usually select u to be the function that becomes simpler.

Answer 2

Int = integral of ...
k and c = contants
exp = exponent

A) Simple integration

1) int dx = x + c

2) Int k dx = k int dx = kx + c

3) Int (x exp n) dx = (x exp n +1) on (n +1)

4) Int (kx + r) dx = Int kx dx + Int r dx = (kx exp2) on 2+ rx + c


B) Integration by parts:

Int udv = uv - Int vdu


C) integration of main Trigonometric functions

1) Int sin x dx = - cos x + c

2) Int cos x dx = sin x + c

3) Int tan x dx = -ln x + c

4) Int cotan x = ln x + c


C) Integration by substitutions

Int sin 5x dx

Put u = 5x
du = 5 dx imply dx = du on 5


Int sin u du = cos u du on 5 + c (cos 5x ) on 5 + c

Answer 3

?sin(5x)cos(7x) dx word that sin(12x) = sin(5x)cos(7x) + cos(5x)sin(7x) and sin(-2x) = sin(5x)cos(7x) - cos(5x)sin(7x) Now including the two equations: sin(12x) - sin(2x) = 2sin(5x)cos(7x) => sin(5x)cos(7x) = a million/2 * [sin(12x) - sin(2x)] a million/2 * ?sin(12x) - sin(2x) dx a million/2 * [-a million/12*cos(12x) + a million/2*cos(2x)] + C

Answer 4

Int(VdU)
V*U + Int(UdV)