integration problem (1 Viewer)

[Steven12]

can somebody explain to me how this works?

integration of cos(x)sin^2(x)dx=sin^3x/3 +c

see i could see why integratin cos(x) would disappear.

i get the part where adding one to the power(of sinx) and dividing it.

 

[CM_Tutor]

I = int cos x * sin²x dx

Let u = sin x. It follows that du = cos x dx

So, I = int sin²x * cos x dx = int u² du = u³ / 3 + C = sin³x / 3 + C, for some constant C

 

[fitz33]

Integral of [f(x)]^n * f'(x) = ([f(x)]^(n+1))/(n+1)
in this case, f(x) = sinx and f'(x) = cosx

 

[:: ck ::]

integral f'(x)f(x)^n = f(x)^n+1 / n+1

 

[Supra]

do we have to know that verbatim

 

[Teoh]

It's a rule of integration...so you can choose to know it if you want

 

[Calculon]

Originally posted by Supra
do we have to know that verbatim

fuck I hope not

 

[Xayma]

Originally posted by Supra
do we have to know that verbatim

Its pretty easy to remember, considering it is pretty much the same form of f'(x)e^f(x)

 

[George W. Bush]

Well, yes.

Although it could be a bit confusing for some when it's written in the general case.

 

[kimmeh]

Originally posted by :: ryan.cck ::
integral f'(x)f(x)^n = f(x)^n+1 / n+1

thats how i remeber it too

Originally posted by Teoh
It's a rule of integration...so you can choose to know it if you want

not really a rule i'd say, but a shortcut

 

[:: ck ::]

yupyup shortcut - instead of using substitution.. through which the derivative will be cancelled out

 

[ToO LaZy ^*]

how do you do this type of question...?

 

[George W. Bush]

Let u = a-x, du=-dx
when x=a, u=0
x=0, u=a
switch the limits with the minus from the du.

part two, cos(pi/2 - x) = sinx
letting I = the integral,
2I = integral 0->pi/2 (sin^3 + cos^3)/(sin^3 + cos^3) = integral 0->pi/2 1

 

[ToO LaZy ^*]

Originally posted by George W. Bush

switch the limits with the minus from the du.

like this...?
= -[integral 0->a (f(u)du)]
= [integral a->0 (f(u)du)]
= LHS???..

 

[George W. Bush]

Yes.

integral 0->a f(x)fx = integral 0->a f(u)fx
proof, if necessary
lhs = F(a) - F(0)
rhs = F(a) - F(0)

 

[ToO LaZy ^*]

Originally posted by George W. Bush


part two, cos(pi/2 - x) = sinx
letting I = the integral,
2I = integral 0->pi/2 (sin^3 + cos^3)/(sin^3 + cos^3) = integral 0->pi/2 1

is that the full working?...coz i don't really understand this bit...could you please explain it

 

[CM_Tutor]

Part One: Show that int (from 0 to a) f(x) dx = int (from 0 to a) f(a - x) dx

Consider I = int (from 0 to a) f(x) dx

Let u = a - x. It follows that du = - dx, and when x = 0, u = a, and when x = a, u = 0.

So, I = int (from a to 0) f(a - u) * -1 du = - int (from a to 0) f(a - u) du = int (from 0 to a) f(a - u) du.

But, u is a dummy variable, and hence can be replaced by any other variable.
That is, int (from 0 to a) f(a - u) du = int (from 0 to a) f(a - x) dx.

Hence, I = int (from 0 to a) f(x) dx = int (from 0 to a) f(a - x) dx, as required.

Part Two: Show that int (from 0 to pi/2) cos³x / (cos³x + sin³x) dx = int (from 0 to pi/2) sin³x / (sin³x + cos³x) dx

Applying the result from part one, it follows that:

LHS = int (from 0 to pi / 2) cos³x / (cos³x + sin³x) dx
= int (from 0 to pi / 2) cos³[(pi / 2) - x] / (cos³[(pi / 2) - x] + sin³[(pi / 2) - x]) dx
= int (from 0 to pi / 2) {cos[(pi / 2) - x]}³ / ({cos[(pi / 2) - x]}³ + {sin[(pi / 2) - x]}³) dx
= int (from 0 to pi / 2) [sin x]³ / ([sin x]³ + [cos x]³) dx, as cos[(pi / 2) - x] = sin x and sin[(pi / 2) - x] = cos x
= int (from 0 to pi / 2) sin³x / (sin³x + cos³x) dx
= RHS

Part Three: Hence evaluate these integrals.

Well, from part three we know that
int (from 0 to pi / 2) cos³x / (cos³x + sin³x) dx = int (from 0 to pi / 2) sin³x / (sin³x + cos³x) dx = I (say).

So, 2I = I + I = int (from 0 to pi / 2) cos³x / (cos³x + sin³x) dx + int (from 0 to pi / 2) sin³x / (sin³x + cos³x) dx
2I = int (from 0 to pi / 2) [cos³x / (cos³x + sin³x)] + [sin³x / (sin³x + cos³x)] dx
= int (from 0 to pi / 2) (cos³x + sin³x) / (sin³x + cos³x) dx
= int (from 0 to pi / 2) 1 dx
= [x] (from 0 to pi / 2)
= (pi / 2) - 0

So, 2I = pi / 2, and hence I = pi / 4

 

[ToO LaZy ^*]

OMG...thanks soooo much