Question Number 167024 by peter frank last updated on 04/Mar/22
$$\int\mathrm{sec}\:\theta\mathrm{tan}\:^{\mathrm{4}} \theta\mathrm{d}\theta \\ $$
Answered by cortano1 last updated on 05/Mar/22
![let t = tan θ I=∫ (t^4 /( (√(1+t^2 )))) = ∫ ((t^3 (t))/( (√(1+t^2 )))) dt IBP { ((u=t^3 ⇒du=3t^2 dt)),((v=∫(1/2) ((d(1+t^2 ))/( (√(1+t^2 )))) = (√(1+t^2 )))) :} I=t^3 (√(1+t^2 ))−∫ 3t^2 (√(1+t^2 )) dt I=t^3 (√(1+t^2 ))−3∫t (t(√(1+t^2 ))) dt I=t^3 (√(1+t^2 )) −3[ (1/3)t (√((1+t^2 )^3 ))−(1/3)∫(1+t^2 )^(3/2) dt ] I= t^3 (√(1+t^2 ))−t (√((1+t^2 )^3 ))+∫ (1+t^2 )^(3/2) dt](https://www.tinkutara.com/question/Q167035.png)
$$\:\mathrm{let}\:\mathrm{t}\:=\:\mathrm{tan}\:\theta \\ $$$$\mathrm{I}=\int\:\frac{\mathrm{t}^{\mathrm{4}} }{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}\:=\:\int\:\frac{\mathrm{t}^{\mathrm{3}} \left(\mathrm{t}\right)}{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}\:\mathrm{dt} \\ $$$$\mathrm{IBP}\:\begin{cases}{\mathrm{u}=\mathrm{t}^{\mathrm{3}} \Rightarrow\mathrm{du}=\mathrm{3t}^{\mathrm{2}} \:\mathrm{dt}}\\{\mathrm{v}=\int\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{d}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}\:=\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}\end{cases} \\ $$$$\mathrm{I}=\mathrm{t}^{\mathrm{3}} \sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }−\int\:\mathrm{3t}^{\mathrm{2}} \:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:\mathrm{dt} \\ $$$$\mathrm{I}=\mathrm{t}^{\mathrm{3}} \:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }−\mathrm{3}\int\mathrm{t}\:\left(\mathrm{t}\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right)\:\mathrm{dt} \\ $$$$\mathrm{I}=\mathrm{t}^{\mathrm{3}} \:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:−\mathrm{3}\left[\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{t}\:\sqrt{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{3}}\int\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \mathrm{dt}\:\right] \\ $$$$\mathrm{I}=\:\mathrm{t}^{\mathrm{3}} \:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }−\mathrm{t}\:\sqrt{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{3}} }+\int\:\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:\mathrm{dt}\: \\ $$