$
\eqalign{
& \int {\frac{1}
{{\sin x}}dx} = \ln \frac{{1 - \cos x}}
{{\left| {\sin x} \right|}}\,\,(1.8) \cr
& \int {\frac{{dx}}
{{a^2 + b^2 x^2 }} = \frac{1}
{{ab}}\arctan \frac{{bx}}
{a}} \,\,\,(1.19) \cr
& \int {\sin mx \cdot \cos nx\,dx = - \frac{{\cos (m - n)x}}
{{2(m - n)}}} - \frac{{\cos (m + n)x}}
{{2(m + n)}}\,\,(m^2 \ne n{}^2)\,\,(1.16) \cr
& \int {\sqrt {a^2 + x^2 } } dx = \frac{x}
{2}\sqrt {a^2 + x^2 } + \frac{{a^2 }}
{2}\ln \left( {x + \sqrt {a^2 + x^2 } } \right)\,\,\left( {a \in {R}^ + } \right)\,\left( {1.25} \right) \cr
& \int {\frac{1}
{{a^2 - b^2 x^2 }}} \,dx = \frac{1}
{{2ab}}\ln \left| {\frac{{a + bx}}
{{a - bx}}} \right|\,\,\,\left( {a,b \in {R}^ + } \right)\,\,\left( {1.20} \right) \cr
& \int {\frac{{dx}}
{{\sqrt {a^2 - x^2 } }}} = \arcsin \left( {\frac{x}
{a}} \right)\,\,\left( {a \in {R}^ + } \right)\left( {1.21} \right) \cr}
$
