Trouble with integral

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Hi, Ive been having trouble with this one its probly easy, but I cant get started.

[tex]\int sinxcos(cosx)[/tex]

I tried integrating by parts,

u=cos(cosx)
du=-sin(cosx)(-sinx)......is this even right?
dv=sinx
v=-cosx
but it looks as if doing that made it alot harder, can anyone point me in the right direction?thanks
 

shmoe

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Did you try a substitution?
 
[tex]u=cosx[/tex]
[tex]du=-sinx[/tex]
[tex]-\int cosu du[/tex]
[tex]=-sin u[/tex]
[tex]=-sin(cosx)[/tex]

I pray this is wrong because if it isnt i'm retarted
 

shmoe

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That's it. (plus a constant, +C)
 
I have another question which isnt explained in the text, when trying to find out A B and C in a partial fraction integral, how do you do it, I read online somewhere about gaussian elimination, it wasnt well explained though, is this how, or is there another way, can you explain?

Thanks
 

HallsofIvy

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Gaussian elimination???

I assume that by "A B and C in a partial fraction integral" you mean, for example, the A, B, C is
[tex]\frac{1}{(x-1)(x-2)(x-3)= \frac{A}{x-1}+ \frac{B}{x-2}+ \frac{C}{x-2}[/tex]

Multiply the entire equation by the denominator (x-1)(x-2)(x-3) and you get
[tex]1= A(x-2)(x-3)+ B(x-1)(x-2)+ C(x-1)(x-2)[/tex]

Now let x= 1, 2, 3 in succession and A, B, C fall out. If you denominators are (x-a)2 or x2+ 1 you might need to let x equal other numbers to get additional equations to solve.
 

shmoe

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There's more than one way to approach it. An example:

[tex]\frac{1}{x^2-1}=\frac{1}{(x-1)(x+1)}[/tex]

The denominator is distinct linear factors, so we know we will have:

[tex]\frac{1}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}[/tex]

for some A and B. this becomes:

[tex]\frac{1}{x^2-1}=\frac{A(x+1)}{x^2-1}+\frac{B(x-1)}{x^2-1}=\frac{Ax+A+Bx-B}{x^2-1}=\frac{(A+B)x+(A-B)}{x^2-1}[/tex]

Next equate the numerators:

[tex]0x+1=(A+B)x+(A-B)[/tex]

the 'placeholder' 0x is so you see the linear term is zero on the left.

Equating coefficients you now have a system of 2 equations and 2 variables:

0=A+B
1=A-B

Gaussian elimination is a general method for solving linear systems like this, but you can use whatever you like. Here, A=-B from the first equation, then the second becomes 1=2*A, so A=1/2 and B=-1/2.

Rather than collect terms when you combined the fractions on the right hand side above, you could equate the numerators as:

[tex]1=A(x+1)+B(x-1)[/tex]

and sub in some convenient values for x. x=1 gives 1=A*(2)+B*(0) so A=1/2. x=-1 gives 1=A*(0)+B*(-2), so B=-1/2.
 
i understand what you are saying..heres what im trying to figure out...

A+B=1
A-B+C=0
4A-6B-3C=0
?
 

shmoe

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Well you can use the first equation to write B in terms of A, giving B=1-A.

Substitute this value of B into the 2nd and 3rd equations and you now have 2 equations with 2 unknowns (A and C). Solve either equation for either A or C and substitute into the remaining equation and you'll have one of the variables. Can you see how to do this and where to go from here?
 
yea i think im getting it...Il see if i can get an answer!
 
ok, that wasnt too tough, i think i got it
 
one final question..if i have [tex] \frac{x^3}{(x+3)^3}[/tex]..would it be this...[tex]\frac{A}{x+3} + \frac{B}{(x+3)^2} + \frac {C}{(x+3)^3}[/tex]..or would i have A B C and D...[tex]\frac{A}{1} + \frac{B}{x+3}[/tex]and so on?
 

0rthodontist

Science Advisor
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In that case you have to do polynomial division first to get the numerator in lower terms than the denominator, and then it's the 3 term decomposition.
 
oooooone more..heres my question..[tex]\int ln(x^2-1)dx[/tex]..i let..
u=ln(x^2+1)
du=2x
dv=dx
v=x
so....i kept going and got [tex]ln(x^2-1)x - \int \frac{2x^2}{x^2-1}dx[/tex]..would this partial fraction be A/x+1 +(B/x-1)?if so im confused because i get A+B=0 and A-B=0
 
Last edited:
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suspenc3 said:
oooooone more..heres my question..[tex]\int ln(x^2-1)dx[/tex]..i let..
u=ln(x^2+1)
du=2x
dv=dx
v=x
so....i kept going and got [tex]ln(x^2-1)x - \int \frac{2x^2}{x^2-1}dx[/tex]..would this partial fraction be A/x+1 +(B/x-1)?if so im confused because i get A+B=0 and A-B=0
If u is ln(x^2+1) then du is definitely not 2x. What is the derivative of the natural log?
 
yeah my bad..i meant to say [tex]2x \frac{1}{x^2-1}[/tex]
 
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suspenc3 said:
oooooone more..heres my question..[tex]\int ln(x^2-1)dx[/tex]..i let..
u=ln(x^2+1)
du=2x
dv=dx
v=x
so....i kept going and got [tex]ln(x^2-1)x - \int \frac{2x^2}{x^2-1}dx[/tex]..would this partial fraction be A/x+1 +(B/x-1)?if so im confused because i get A+B=0 and A-B=0
Before you try decomposing into partial fractions you should divide 2x2 by x2-1 and that should give you something pluse a linear term over x2-1.
 
good thinkin, I thinkI got it, THanks leet
 
1,072
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suspenc3 said:
good thinkin, I thinkI got it, THanks leet
Your welcome, glad I could help.
 

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