# P5: Reduction Formulae

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#1

In = INT sec^n x dx

(n-1)In = tanx sec^(n-2)x + (n-2)In-2

Just a hint on how to start would be great, I'm at a loss.
0
15 years ago
#2
Integrate by parts, choose u = sec^(n-2)x and dv/dx = sec^2(x).
0
#3
Thanks 0
15 years ago
#4
(Original post by queen-of-pain)

In = INT sec^n x dx

(n-1)In = tanx sec^(n-2)x + (n-2)In-2

Just a hint on how to start would be great, I'm at a loss.

I've only done up to P3 so this may be wrong, but I'm guessing:

Int.sec^nx.dx=int.sec^2x.sec^(n-2)x.dx

let u=sec^(n-2)x v'=sec^2x
u'=(n-2)sec^(n-2)x.tanx v=tanx

so int.sec^nx= sec^(n-2)xtanx-int.(n-2)sec^(n-2)xtan^2x.dx
= sec^(n-2)xtanx-int.(n-2)sec^(n-2)x(sec^2x-1).dx
= sec^(n-2)xtanx-(n-2)int.sec^nx+(n-2)int.sec^(n-2)x.dx
(n-1)int.sec^nx.dx=tanx.sec^(n-2)x+(n-2)int.sec^(n-2)x.dx

which I think is the same thing.
0
15 years ago
#5
(Original post by Ralfskini)
I've only done up to P3 so this may be wrong, but I'm guessing:

Int.sec^nx.dx=int.sec^2x.sec^(n-2)x.dx

let u=sec^(n-2)x v'=sec^2x
u'=(n-2)sec^(n-2)x.tanx v=tanx

so int.sec^nx= sec^(n-2)xtanx-int.(n-2)sec^(n-2)xtan^2x.dx
= sec^(n-2)xtanx-int.(n-2)sec^(n-2)x(sec^2x-1).dx
= sec^(n-2)xtanx-(n-2)int.sec^nx+(n-2)int.sec^(n-2)x.dx
(n-1)int.sec^nx.dx=tanx.sec^(n-2)x+(n-2)int.sec^(n-2)x.dx

which I think is the same thing.
Excuse me, can anyone tell me whether the followig is acceptable:
Int.sec^nx.dx
=int.sec^2x.sec^(n-2)x.dx
=int.sec^(n-2)x d(sec^2x) (<---- is it allowed?)
sec^(n-2)xtanx-int.(n-2)sec^(n-2)xtan^2x.dx
0
15 years ago
#6
Yes it is acceptable but in this case it is wrong. We have

INT sec^nx dx
= INT sec^(n-2)x (sec^2x dx)

Now, you know that

d(tanx)/dx = sec^2x

Therefore,

d(tanx) = sec^2x dx

and then

INT sec^(n-2)x (sec^2x dx) = INT sec^(n-2)x d(tanx)
So, it is justified and I have seen it used once or twice in substitutions and when using parts. For example, say you have an integral with respect to x: INT f(x) dx and you apply some substitution, say u=x^2 which gives you an integral in some form INT g(u) du. You should remember that what you are actually doing now is integrating with repect to x^2 becuase u=x^2; that is, INT g(u) du = INT g(u) d(x^2). Also, for integration by parts we have

INT u dv/dx dx = uv - INT v du/dx dx

Which can actually be written

INT u dv = uv - INT v du

So, for example

INT xe^x dx = INT x d(e^x)

Comparing this to the above formula with u = x and v = e^x gives

INT x d(e^x) = xe^x - INT e^x d(x) = (x-1)e^x + C

So, the step is a valid one but I personally do not consider it to be a particularly practical or neat method.
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