Harder Integration methods (1 Viewer)

[rsingh]

Hey guys, I've seen some integration questions, especially on James Ruse past papers, where the integration questions require further knowledge than just the basic formulas. Like different formuals for exponentials, and integration by substituion.

Anyone have any idea what I'm talking about?
For example, this one is new to me: /f'(x)f(x)^ndx = f(x)^n + c.
I've never seen that one before, but I've found it quiet useful in 3u.

If someone could post up these and other methods of integration, that would be really helpful!

Thanks.

 

[FinalFantasy]

Hey guys, I've seen some integration questions, especially on James Ruse past papers, where the integration questions require further knowledge than just the basic formulas. Like different formuals for exponentials, and integration by substituion.

Anyone have any idea what I'm talking about?
For example, this one is new to me: /f'(x)f(x)^ndx = f(x)^n + c.
I've never seen that one before, but I've found it quiet useful in 3u.

If someone could post up these and other methods of integration, that would be really helpful!

Thanks.

int. f(x)^nf'(x) dx=int. f(x)^n d(f(x))=(f(x))^(n+1) \(n+1) +C
juz do dat, no need that formula

forgot the +1 at the end

 

[rama_v]

Yeah I've noticed in teh Ruse papers they also do other hard integration methods that are (I think) taught in 4 unit but are also applicable for 3 unit students (though I doubt they will ever put them in the hsc)

Things like splitting up a function using polynomial division, and also dividing both numerator and denominator by some variable, etc.

 

[Slidey]

Uh. Yes you have. It's a specific application of the chain rule.

Oh, and you mean Int[n.f'(x)f(x)^(n-1)]dx = f(x)^n + C btw.
Since d(f(x)^n)/dx=n.f'(x).f(x)^(n-1)

If you can't do it in your head, use the substitution:
u=f(x)
du=f'(x) dx
Thus the integral becomes: Int(n.u^(n-1).du) = n.u^n/n + C = f(x)^n + c

 

[Slidey]

Here's a trick you should all know:
First, acknowledge that Int(f(x))dx from a to b is the same as Int(f(u))du from a to b.
Now, then, prove that:
Int(f(x))dx from 0 to a = Int(f(a-x))dx from 0 to a

Answer:
Use the substitution x=a-u
dx=-du
Terminals: 0 becomes a, a becomes 0:
Int(f(x))dx from 0 to a
Becomes
-Int(f(a-u))du from a to 0
Now switch terminals and remove the negative sign, it becomes:
Int(f(a-u))du from 0 to a

Thus Int(f(x))dx from 0 to a = Int(f(a-x))dx from 0 to a (see second line of post if unsure)

 

[Slidey]

Using the above post, evaluate:

Int(x(1-x)^5) from 0 to 1
And:
Int(cos^2(x)) from 0 to pi/2

Also, using the substitution x=pi-y, prove that:
Int(xsin^3(x))dx from 0 to pi = 2pi/3

And another:
Prove that:
Int(xsinx/(1+cos^2(x))dx from 0 to pi = Int((pi-x)sinx/(1+cos^2(x))dx from 0 to pi
=pi^2/4

EDIT: Eh, wait. This is a 3unit thread. My apologies. Still, try them.

 

[rsingh]

Sorry guys, I got the formula wrong.
It's int. f'(x)f(x)^ndx = f(x)^(n+1)/(n+1) + C.

So do we ever need to use long division for some integrals.
For example, how do you integrate?

int. x^2 + 1 / (x^2 + 4) dx ?

 

[FinalFantasy]

Sorry guys, I got the formula wrong.
It's int. f'(x)f(x)^ndx = f(x)^n/(n+1) + C.

So do we ever need to use long division for some integrals.
For example, how do you integrate?

int. x^2 + 1 / (x^2 + 4) dx ?

int. (x²+1)\(x²+4) dx u can use long division
but y not try dis..
int. (x²+1)\(x²+4) dx=int. (x²+4-3)\(x²+4) dx
=int. (x²+4)\(x²+4) dx-int. 3\(x²+4) dx

 

[rsingh]

hmm, I think I follow.
However, how would you integrate from there?
I mean, woudn't you have to use long divison to do int. 3/(x^2+4) dx.

 

[Slidey]

int 3/(x^2+4)

think of inverse tan.

3/2*inverse tan(x/2)

 

[FinalFantasy]

integrate 3\(x²+4) dx is inverse tan
look on ur formula sheet,
integrate 3\(x²+4) dx=3*(1\2) tan^-1 (x\2)

 

[rsingh]

Oh that's right. Thanks.
We haven't completed inverse trig. - it's our last topic.
But if you could give me a rundown on how to differentiate and integrate inverse trig functions, that would be greaT!

 

[Slidey]

arcsin(x/a) -> derivative is 1/sqrt(a^2-x^2)
arccos(x/a) -> derivative is -1/sqrt(a^2-x^2) (so similar to sine you can ignore it and simply remember it is negative derivative of sine)
arctan(x/a) -> derivative is a/(a^2+x^2)

 

[velox]

How about reading the maths textbook?

 

[velox]

(so similar to sine you can ignore it and simply remember it is negative derivative of sine)

Which is frowned upon.

 

[rsingh]

Is it where a is a constant, or a pronumeral?

Yeah velox, that would be a great idea, if I could make some sense of it - haha.

So when integrating you just to the reverse?
Oh yes, and I don't know what you mean by arcsin(x/a), etc?

 

[Slidey]

Which is frowned upon.

d(arcsinx)/dx=-d(arccosx)/dx is completely valid. I would love to see you disprove it.

 

[LaCe]

U guys should already know that, it is nothing new.

I guess 4u helps ur 3u alot
like in Integration, Graph Sketching, and Volumes
And of course the Harder 3u Section although we do not cover that is class, its just harder 3u questions (not just hard but almost fucking impossible)

 

[Slidey]

Is it where a is a constant, or a pronumeral?

Yeah velox, that would be a great idea, if I could make some sense of it - haha.

So when integrating you just to the reverse?
Oh yes, and I don't know what you mean by arcsin(x/a), etc?

a is a constant, like 1 or 4.

Yes, integrating goes backwards. Integration is the inverse operation of differentiation.

arc stands for inverse. arcsin(x)=sin^-1(x)

 

[velox]

When did i say i'd disprove it? And i was talking about say if you had - 1/sqrt(1-x^2) and you had to integrate it and your answer was -arcsin then it would be frowned upon.