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

[suspenc3]

Hi, I've been having trouble with this one its probly easy, but I can't get started.

\int sinxcos(cosx)

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

 

[shmoe]

Did you try a substitution?

 

[suspenc3]

u=cosx
du=-sinx
-\int cosu du
=-sin u
=-sin(cosx)

I pray this is wrong because if it isn't I'm retarted

 

[shmoe]

That's it. (plus a constant, +C)

 

[suspenc3]

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

 

[HallsofIvy]

Gaussian elimination?

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

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

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]

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

\frac{1}{x^2-1}=\frac{1}{(x-1)(x+1)}

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

\frac{1}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}

for some A and B. this becomes:

\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}

Next equate the numerators:

0x+1=(A+B)x+(A-B)

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:

1=A(x+1)+B(x-1)

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.

 

[suspenc3]

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
?

 

[shmoe]

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?

 

[suspenc3]

yea i think I am getting it...Il see if i can get an answer!

 

[suspenc3]

ok, that wasnt too tough, i think i got it

 

[suspenc3]

one final question..if i have \frac{x^3}{(x+3)^3}..would it be this...\frac{A}{x+3} + \frac{B}{(x+3)^2} + \frac {C}{(x+3)^3}..or would i have A B C and D...\frac{A}{1} + \frac{B}{x+3}and so on?

 

[0rthodontist]

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.

 

[suspenc3]

oooooone more..heres my question..\int ln(x^2-1)dx..i let..
u=ln(x^2+1)
du=2x
dv=dx
v=x
so...i kept going and got ln(x^2-1)x - \int \frac{2x^2}{x^2-1}dx..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

 

[d_leet]

oooooone more..heres my question..\int ln(x^2-1)dx..i let..
u=ln(x^2+1)
du=2x
dv=dx
v=x
so...i kept going and got ln(x^2-1)x - \int \frac{2x^2}{x^2-1}dx..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?

 

[suspenc3]

yeah my bad..i meant to say 2x \frac{1}{x^2-1}

 

[d_leet]

oooooone more..heres my question..\int ln(x^2-1)dx..i let..
u=ln(x^2+1)
du=2x
dv=dx
v=x
so...i kept going and got ln(x^2-1)x - \int \frac{2x^2}{x^2-1}dx..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.

 

[suspenc3]

good thinkin, I thinkI got it, THanks leet

 

[d_leet]

good thinkin, I thinkI got it, THanks leet

Your welcome, glad I could help.