Question

Question #e2f56

Answer 1

`tan(x/2)+C`

Explanation:

`cos^2(x/2) = (1+cos(x))/2`
So `1+cos(x) = 2cos^2(x/2)` and `1/(1+cos x) =1/2sec^2(x/2)`
`int\ 1/2sec^2(x/2)\ dx = 1/2int\ sec^2(x/2)\ dx`

Substituting `u=x/2` and `du=dx/2`, we get `1/2int\ 2sec^2(u)\ du=tan(u)+C=tan(x/2)+C`