1) expand in partial fractions.
2) Using cos(2x) = 2 cos^2(x) -1 derive a formula for cos(1/2 x) in terms of cos(x). Express the integrand in terms of cos(1/2 x)
3) [Notation: cos = c, sin = s]
1/c^4 = (s^2 + c^2)/c^4 =
1/c^2 + s^2/c^4
1/c^2 yields a tangent when integrated
To integrate s^2/c^4 do partial integration:
s^2/c^4 = s (s/c^4)
Integral of s/c^4 is 1/3 1/c^3
So, we need to integrate
1/c^3
times the derivative of of s, i.e. 1/c^2, but that is tan(x)!
Note that the way to solve such problems is not to systematically work things out in detail at first, because then you would take too long to see that a method doesn't work.
Instead, you should reason like I just did, i.e. forget the details, be very sloppy, just to see if things works out and you get an answer in principle, even though you need to fill in the details.
If you get better at this, you can do the selection of what method to use in your head, you'll see it in just a few seconds when looking at an integral. You can then start to work out the solution for real on paper.
I have 3 questions, and I cannot find method that actually solves them.
1) Integral [(4s+4)/([s^2+1]*([S-1]^3))]
2) Integral [ 2*sqrt[(1+cosx)/2]]
3) Integral [ 20*(sec(x))^4
Thanks in advance.
3 answers
Could you expand on #1 a bit more? I tried Partial Fractions, but I couldn't get a definite answer...
I'll show you how to do it without solving any equations.
The function is up to a factor 4:
f(s) = (s+1)/{(s^2+1)*[(s-1)^3]}
The partial fractions are precisely the singular terms when expanding around the singularities. So, let's examine the singularity at s = 1:
Put s = 1 + u and expand in powers of u, keeping only the singular terms:
f(1+u) = u^(-3) (2+u)/[(u+1)^2+1] =
u^(-3) (2+u)/(2+2u+u^2) =
u^(-3) (2+u)/2 1/(1+u+u^2/2) = (use
1/(1+x) = 1-x+x^2-x^3+...)
u^(-3) (2+u)/2 [1-u-u^2/2 +u^2+...]
=[u^(-3) + u^(-2)/2][1-u+u^2/2+...]
u^(-3) -1/2 u^(-2) + nonsingular terms
This means that in the neighborhood of s = 1 we have:
f(s) = 1/(s-1)^3 - 1/2 1/(s-1)^2 +.. nonsingular terms
f(s) also has singularities at s = i and s =-i. Around s = i, we have:
f(s) = a/(s-i) + nonsingular terms
Multiply both sides by (s-i) and take the limit s --> i to find a:
a = (i+1)/[2i(i-1)^3] =
-1/2 1/(i-1)^2
Around s = -i, we have:
f(s) = b/(s+i) + nonsingular terms
Multiply both sides by (s+i) and take the limit s --> -i to find b:
b = (-i+1)/[-2i(-i-1)^3] =
-1/2 1/[(i+1)^2]
The sum of all the singular terms of the expansions around the singular points is:
1/(s-1)^3 - 1/2 1/(s-1)^2 +
-1/2 1/(i-1)^2 1/(s-i)
-1/2 1/[(i+1)^2] 1/(s+i) =
1/(s-1)^3 - 1/2 1/(s-1)^2 +
1/2 1/(s^2+1)
This is the desired partial fraction decomposition. The proof of why this works is a consequence of Liouville's theorem
http://en.wikipedia.org/wiki/Liouville's_theorem_(complex_analysis)#Proof
Just consider the difference of f(s) and the sum of all the singular terms. The resulting function doesn't have any singularites and is bounded. So, by Liouville's theorem it is a constant. We know that f(s) and all the singular terms tend to zero, so that constant must be zero.
The function is up to a factor 4:
f(s) = (s+1)/{(s^2+1)*[(s-1)^3]}
The partial fractions are precisely the singular terms when expanding around the singularities. So, let's examine the singularity at s = 1:
Put s = 1 + u and expand in powers of u, keeping only the singular terms:
f(1+u) = u^(-3) (2+u)/[(u+1)^2+1] =
u^(-3) (2+u)/(2+2u+u^2) =
u^(-3) (2+u)/2 1/(1+u+u^2/2) = (use
1/(1+x) = 1-x+x^2-x^3+...)
u^(-3) (2+u)/2 [1-u-u^2/2 +u^2+...]
=[u^(-3) + u^(-2)/2][1-u+u^2/2+...]
u^(-3) -1/2 u^(-2) + nonsingular terms
This means that in the neighborhood of s = 1 we have:
f(s) = 1/(s-1)^3 - 1/2 1/(s-1)^2 +.. nonsingular terms
f(s) also has singularities at s = i and s =-i. Around s = i, we have:
f(s) = a/(s-i) + nonsingular terms
Multiply both sides by (s-i) and take the limit s --> i to find a:
a = (i+1)/[2i(i-1)^3] =
-1/2 1/(i-1)^2
Around s = -i, we have:
f(s) = b/(s+i) + nonsingular terms
Multiply both sides by (s+i) and take the limit s --> -i to find b:
b = (-i+1)/[-2i(-i-1)^3] =
-1/2 1/[(i+1)^2]
The sum of all the singular terms of the expansions around the singular points is:
1/(s-1)^3 - 1/2 1/(s-1)^2 +
-1/2 1/(i-1)^2 1/(s-i)
-1/2 1/[(i+1)^2] 1/(s+i) =
1/(s-1)^3 - 1/2 1/(s-1)^2 +
1/2 1/(s^2+1)
This is the desired partial fraction decomposition. The proof of why this works is a consequence of Liouville's theorem
http://en.wikipedia.org/wiki/Liouville's_theorem_(complex_analysis)#Proof
Just consider the difference of f(s) and the sum of all the singular terms. The resulting function doesn't have any singularites and is bounded. So, by Liouville's theorem it is a constant. We know that f(s) and all the singular terms tend to zero, so that constant must be zero.