Solving Integral Troubles: Tips and Tricks for Integrating Sin and Cos Functions

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In summary, the conversation involves a person having trouble with integrating a complex function and seeking help from others. They try different methods and discuss the use of Gaussian elimination to solve for variables in a partial fraction integral. They also discuss polynomial division and using it to simplify the function before decomposing into partial fractions. The conversation ends with the person thanking the others for their help.
  • #1
suspenc3
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Hi, I've been having trouble with this one its probly easy, but I can't 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 a lot harder, can anyone point me in the right direction?thanks
 
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  • #2
Did you try a substitution?
 
  • #3
[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 isn't I'm retarted
 
  • #4
That's it. (plus a constant, +C)
 
  • #5
I have another question which isn't 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
 
  • #6
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.
 
  • #7
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.
 
  • #8
i understand what you are saying..heres what I am trying to figure out...

A+B=1
A-B+C=0
4A-6B-3C=0
?
 
  • #9
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?
 
  • #10
yea i think I am getting it...Il see if i can get an answer!
 
  • #11
ok, that wasnt too tough, i think i got it
 
  • #12
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?
 
  • #13
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.
 
  • #14
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 I am confused because i get A+B=0 and A-B=0
 
Last edited:
  • #15
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 I am 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?
 
  • #16
yeah my bad..i meant to say [tex]2x \frac{1}{x^2-1}[/tex]
 
  • #17
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 I am 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.
 
  • #18
good thinkin, I thinkI got it, THanks leet
 
  • #19
suspenc3 said:
good thinkin, I thinkI got it, THanks leet

Your welcome, glad I could help.
 

What is the purpose of integrating sin and cos functions?

The purpose of integrating sin and cos functions is to find the area under the curve of these functions. This can be useful in solving real-world problems involving periodic motion or in finding the total displacement or distance traveled.

What are the basic rules for integrating sin and cos functions?

The basic rules for integrating sin and cos functions include using the power rule, the product rule, and the chain rule. Additionally, the trigonometric identities, such as the double angle formula and the half angle formula, can also be used to simplify the integration process.

How do I solve complex integrals involving sin and cos functions?

To solve complex integrals involving sin and cos functions, it is important to break down the integral into smaller parts using the trigonometric identities and then apply the basic rules of integration. It can also be helpful to use substitution or integration by parts to simplify the integral.

What are some common mistakes to avoid when integrating sin and cos functions?

Some common mistakes to avoid when integrating sin and cos functions include forgetting to use the correct trigonometric identities, not simplifying the integral using algebraic manipulation, and making errors in calculation. It is also important to pay attention to the limits of integration and to check for any discontinuities in the function.

Can I use technology to help me solve integrals involving sin and cos functions?

Yes, technology such as graphing calculators or online integral calculators can be used to help solve integrals involving sin and cos functions. However, it is important to understand the basic rules and techniques for integrating these functions in order to use technology effectively and to check the results for accuracy.

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