integrate f(x)=square root of 1+(cosx)^2?

Question

please show working

Answer 1

even my voyage refuses to give an answer but here is the link that should help you...

∫ √(1 + cos²(x)) dx is an irraional function... use the very first formula in the list of integrals from the page I have provided *** source...

x = 1 and a = cos(x)

use this formula...

http://img204.imageshack.us/my.php?image=d6a05beb22d8183fb485643nr9.png

ur final answer will be...

1/2 { 1 * √(1 + cos²(x)) + cos²(x) * ln[1 + √(1 + cos²(x))]}

Answer 2

3+3=6

Answer 3

Integral (sqrt [1 + (cosx)^2]) dx

Use this identity

cos^2(x) = (1 + cos2x)/2. You should get

Integral (sqrt [1 + [(1 + cos2x)/2]])dx

Make one fraction inside the square root.

Integral (sqrt ([2 + 1 + cos2x]/2))dx

Integral (sqrt [3 + cos2x]/2)dx

We can take square root of top and bottom, allowing us to pull 1/sqrt(2) out of the integral.

1/sqrt(2) * Integral (sqrt [3 + cos2x])dx

I seem to be stuck here. This may not be solveable using the methods we know.

Answer 4

This is a complete elliptic integral of the second kind:
Sqrt[2] E[x,(1\2)].
Look it up in the tables or use SW to evaluate it numerically.

Answer 5

I'll do the 2nd one. Use x = sinu substitution. dx/du = cosu dx = cosu du ∫ dx/(1-x²)³/² = ∫ (cosu du)/(1-sin²u)³/² = ∫ (cosu du)/(cos²u)³/² = ∫ (cosu du)/(cos³u) = ∫ du/(cos²u) = ∫ sec²u du = tanu + C Since sinu = x, cosu = √(1-x²), and tanu = sinu/cosu, giving us ∫ dx/(1-x²)³/² = x/√(1-x²) + C

Answer 6

1+-2sinx