## Thomas' Calculus 13th Edition

$$- \ln 3$$
\eqalign{ & \int_0^{\pi /3} {\frac{{4\sin \theta }}{{1 - 4\cos \theta }}} d\theta \cr & {\text{Use substitution:}}\cr & {\text{Let }}u = 1 - 4\cos \theta,{\text{ so that }}du = 2\sin \theta d\theta \cr & {\text{The new limits on }}u{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}\theta = \pi /3,{\text{ }}u = 1 - 4\cos \left( {\pi /3} \right) = - 1 \cr & \,\,\,\,\,\,{\text{If }}\theta = 0,{\text{ }}u = 1 - 4\cos \left( 0 \right) = - 3 \cr & {\text{write the integral in terms of }}u \cr & \int_0^{\pi /3} {\frac{{4\sin \theta }}{{1 - 4\cos \theta }}} d\theta = \int_{ - 3}^{ - 1} {\frac{{du}}{u}} \cr & {\text{integrate}} \cr & = \left( {\ln \left| u \right|} \right)_{ - 3}^{ - 1} \cr & {\text{use the fundamental theorem of calculus: }}\cr & \int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr & = \ln \left| { - 1} \right| - \ln \left| { - 3} \right| \cr & = - \ln 3 \cr}