## Thomas' Calculus 13th Edition

$$\ln 2$$
\eqalign{ & \int_0^{\pi /2} {\tan \frac{x}{2}} dx \cr & {\text{Use substitution:}}\cr & {\text{Let }}u = \frac{x}{2},{\text{ so that }}du = \frac{1}{2}dx \cr & {\text{The new limits on }}u{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}x = \pi /2,{\text{ }}u = \frac{{\pi /2}}{2} = \pi /4 \cr & \,\,\,\,\,\,{\text{If }}x = 0,{\text{ }}u = \frac{0}{2} = 0 \cr & {\text{Write the integral in terms of }}u \cr & \int_0^{\pi /2} {\tan \frac{x}{2}} dx = \int_0^{\pi /4} {\tan u} \left( {2du} \right) \cr & = 2\int_0^{\pi /4} {\tan u} du \cr & {\text{Integrate }} \cr & = 2\left( { - \ln \left| {\cos u} \right|} \right)_0^{\pi /4} \cr & {\text{Use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr & = - 2\left( {\ln \left| {\cos \frac{\pi }{4}} \right| - \ln \left| {\cos 0} \right|} \right) \cr & {\text{Simplifying, we get:}} \cr & = - 2\left( {\ln \left( {\frac{{\sqrt 2 }}{2}} \right) - \ln \left( 1 \right)} \right) \cr & = - 2\ln \left( {\frac{{\sqrt 2 }}{2}} \right) \cr & = \ln 2 \cr}