The Notes: `int (1 - tan^2(x))/(sec^2(x)) dx` Evaluate the integral

$\displaystyle\int\frac{{{1}-{{\tan}}^{{2}}{\left({x}\right)}}}{{{{\sec}}^{{2}}{\left({x}\right)}}}{\left.{d}{x}\right.}$

Rewrite the integrand as,

$\displaystyle\int\frac{{{1}-{{\tan}}^{{2}}{\left({x}\right)}}}{{{{\sec}}^{{2}}{\left({x}\right)}}}{\left.{d}{x}\right.}=\int\frac{{{1}-\frac{{{{\sin}}^{{2}}{\left({x}\right)}}}{{{{\cos}}^{{2}}{\left({x}\right)}}}}}{{\frac{{1}}{{{{\cos}}^{{2}}{\left({x}\right)}}}}}{\left.{d}{x}\right.}$

$\displaystyle=\int{\left({{\cos}}^{{2}}{\left({x}\right)}-{{\sin}}^{{2}}{\left({x}\right)}\right)}{\left.{d}{x}\right.}$

Now use the following identities:

$\displaystyle{{\cos}}^{{2}}{\left({x}\right)}=\frac{{{1}+{\cos{{\left({2}{x}\right)}}}}}{{2}}$

$\displaystyle{{\sin}}^{{2}}{\left({x}\right)}=\frac{{{1}-{\cos{{\left({2}{x}\right)}}}}}{{2}}$

$\displaystyle=\int{\left(\frac{{{1}+{\cos{{\left({2}{x}\right)}}}}}{{2}}-\frac{{{1}-{\cos{{\left({2}{x}\right)}}}}}{{2}}\right)}{\left.{d}{x}\right.}$