Question

Evaluate the integral\(\int {{\rm{ta}}{{\rm{n}}^{\rm{5}}}{\rm{\theta se}}{{\rm{c}}^{\rm{3}}}{\rm{\theta d\theta }}} \)

Short answer

The integral value of the given equation is\(\int {{\rm{ta}}{{\rm{n}}^{\rm{5}}}{\rm{\theta se}}{{\rm{c}}^{\rm{3}}}{\rm{\theta d\theta }}} {\rm{ = }}\frac{{\rm{1}}}{{\rm{7}}}{\rm{se}}{{\rm{c}}^{\rm{7}}}{\rm{\theta - }}\frac{{\rm{2}}}{{\rm{5}}}{\rm{se}}{{\rm{c}}^{\rm{5}}}{\rm{\theta + }}\frac{{\rm{1}}}{{\rm{3}}}{\rm{se}}{{\rm{c}}^{\rm{3}}}{\rm{\theta + C}}\).

Step-by-step solution

Expand the equation.

Known value\({\rm{1 + ta}}{{\rm{n}}^{\rm{2}}}{\rm{\theta = se}}{{\rm{c}}^{\rm{2}}}{\rm{\theta }}\)and \({\rm{(sec\theta )' = sec\theta tan\theta }}\)

\(\begin{aligned}{c}\int {{\rm{ta}}{{\rm{n}}^{\rm{5}}}} {\rm{\theta se}}{{\rm{c}}^{\rm{3}}}{\rm{\theta d\theta = }}\int {{{\left( {{\rm{ta}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }}} \right)}^{\rm{2}}}} {\rm{se}}{{\rm{c}}^{\rm{2}}}{\rm{\theta sec\theta tan\theta d\theta }}\\{\rm{ = }}\int {{{\left( {{\rm{se}}{{\rm{c}}^{\rm{2}}}{\rm{\theta - 1}}} \right)}^{\rm{2}}}} {\rm{se}}{{\rm{c}}^{\rm{2}}}{\rm{\theta sec\theta tan\theta d\theta }}\end{aligned}\)

Evaluate the equation.

\(\begin{aligned}{c}{\rm{ = }}\left( {\begin{aligned}{*{20}{c}}{{\rm{u = sec\theta }}}\\{{\rm{du = sec\theta tan\theta d\theta }}}\end{aligned}} \right)\\{\rm{ = }}\int {{{\left( {{{\rm{u}}^{\rm{2}}}{\rm{ - 1}}} \right)}^{\rm{2}}}} {{\rm{u}}^{\rm{2}}}{\rm{\;du}}\\{\rm{ = }}\int {\left( {{{\rm{u}}^{\rm{4}}}{\rm{ - 2}}{{\rm{u}}^{\rm{2}}}{\rm{ + 1}}} \right)} {{\rm{u}}^{\rm{2}}}{\rm{\;du}}\\{\rm{ = }}\int {\left( {{{\rm{u}}^{\rm{6}}}{\rm{ - 2}}{{\rm{u}}^{\rm{4}}}{\rm{ + }}{{\rm{u}}^{\rm{2}}}} \right)} {\rm{du}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{7}}}{{\rm{u}}^{\rm{7}}}{\rm{ - }}\frac{{\rm{2}}}{{\rm{5}}}{{\rm{u}}^{\rm{5}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{3}}}{{\rm{u}}^{\rm{3}}}{\rm{ + C}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{7}}}{\rm{se}}{{\rm{c}}^{\rm{7}}}{\rm{\theta - }}\frac{{\rm{2}}}{{\rm{5}}}{\rm{se}}{{\rm{c}}^{\rm{5}}}{\rm{\theta + }}\frac{{\rm{1}}}{{\rm{3}}}{\rm{se}}{{\rm{c}}^{\rm{3}}}{\rm{\theta + C}}\end{aligned}\)

Therefore, the integral value of the given equation is\(\int {{\rm{ta}}{{\rm{n}}^{\rm{5}}}{\rm{\theta se}}{{\rm{c}}^{\rm{3}}}{\rm{\theta d\theta }}} {\rm{ = }}\frac{{\rm{1}}}{{\rm{7}}}{\rm{se}}{{\rm{c}}^{\rm{7}}}{\rm{\theta - }}\frac{{\rm{2}}}{{\rm{5}}}{\rm{se}}{{\rm{c}}^{\rm{5}}}{\rm{\theta + }}\frac{{\rm{1}}}{{\rm{3}}}{\rm{se}}{{\rm{c}}^{\rm{3}}}{\rm{\theta + C}}\).