Saturday

December 20, 2014

December 20, 2014

Posted by **David** on Sunday, March 23, 2008 at 12:12pm.

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.

- Calculus - Integrals -
**Count Iblis**, Sunday, March 23, 2008 at 3:33pm1) 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.

- Calculus - Integrals -
**David**, Sunday, March 23, 2008 at 5:35pmCould you expand on #1 a bit more? I tried Partial Fractions, but I couldn't get a definite answer...

- Calculus - Integrals -
**Count Iblis**, Sunday, March 23, 2008 at 7:43pmI'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.

- Calculus - Integrals -

**Answer this Question**

**Related Questions**

Calculus - Integrals - I have 3 questions, and I cannot find method that ...

Calculus - Find the volume of the solid whose base is the region in the xy-plane...

Calculus - I have two questions, because I'm preparing for a math test on monday...

calc asap! - can you help me get started on this integral by parts? 4 S sqrt(t) ...

Calculus - Use the symmetry of the graphs of the sine and cosine functions as an...

Calc 2 - a. Integral (x^2)/(sqrt(1+(x^2))) Would I separate these two into 2 ...

Calc - Evaluate the integral using any method: (Integral)sec^3x/tanx dx I ...

CALCULUS 2!!! PLEASE HELP!! - I'm having trouble with this question on arc ...

calc - find integral using table of integrals ) integral sin^4xdx this the ...

calc - how do you start this problem: integral of xe^(-2x) There are two ways: 1...