Z-Substitution and Integration

[rheighton]

So the problem is as follows:

1. Show that:

cos(x) = (1-tan^2(x/2)) / (1+tan^2(x/2))

and

sin(x) = (2tan(x/2)) / (1+tan^2(x/2))


2. Using the substitution z = tan(x/2) and the formulas in 1. evaluate:

∫(from 0 to pi/2) dx / (2+cos(x))


I was able to prove the formulas in 1. with some help, but I can't figure out the second part

 

[galactus]

Since you have the identities proofs kicked, just sub the first one in place of cos(x) in your function to integrate.
Then use u...er....z-substitution.

\(\displaystyle \L\\\int_{0}^{\frac{{\pi}}{2}}\frac{dx}{2+cos(x)}\)

\(\displaystyle \L\\\int_{0}^{\frac{\pi}{2}}\frac{dx}{2+\frac{1-tan^{2}{\frac{x}{2}}}{1+tan^{2}{\frac{x}{2}}}}\)

Now, let \(\displaystyle z=tan(\frac{x}{2})\), \(\displaystyle x=2tan^{-1}(z)\)

\(\displaystyle dx=\frac{2}{1+z^{2}}dz\)

This changes the limits of integration to z=0 and z=1 because

\(\displaystyle tan(\frac{\pi}{4})=1\) and \(\displaystyle tan(0)=0\)

\(\displaystyle \L\\\int_{0}^{1}\frac{\frac{2}{1+z^{2}}}{2+\frac{1-z^{2}}{1+z^{2}}}dz\)

=\(\displaystyle \L\\\int_{0}^{1}\frac{2}{z^{2}+3}dz\)

Now, can you finish?.