∫ t sec 2 ( 2 t ) d t {\displaystyle \int t\sec ^{2}\left(2t\right)dt}
u = t d v = sec 2 ( 2 t ) {\displaystyle u=t\qquad dv=\sec ^{2}\left(2t\right)} d u = d t v = 1 2 tan ( 2 t ) {\displaystyle du=dt\qquad v={\frac {1}{2}}\tan \left(2t\right)}
∫ t sec 2 ( 2 t ) d t = 1 2 tan ( 2 t ) − 1 2 ∫ tan ( 2 t ) d t = 1 2 tan ( 2 t ) − 1 4 ∫ tan ( u ) d u = 1 2 tan ( 2 t ) − 1 4 ln | sec 2 t | + c {\displaystyle \int t\sec ^{2}\left(2t\right)dt={\frac {1}{2}}\tan \left(2t\right)-{\frac {1}{2}}\int \tan \left(2t\right)dt={\frac {1}{2}}\tan \left(2t\right)-{\frac {1}{4}}\int \tan \left(u\right)du={\frac {1}{2}}\tan \left(2t\right)-{\frac {1}{4}}\ln |\sec 2t|+c}
u = 2 t d u = 2 d t d u 2 = d t {\displaystyle {\begin{aligned}&u=2t\\[2ex]&du=2dt\\[2ex]&{\frac {du}{2}}=dt\\[2ex]\end{aligned}}}
![{\displaystyle {\begin{aligned}&u=2t\\[2ex]&du=2dt\\[2ex]&{\frac {du}{2}}=dt\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/dbf9adf2af3fdc163b28995dd5fa938699dace0c)