Интеграл tg7x в квадрате как вычеслить

Решите задачу:

$\int\tan^{2}(7x)dx=\int\tan^{2}(7x)\frac{d(7x)}{7}=\left|7x=t;\ d(7x)=dt\right|=\\ =\frac17\int\tan^2tdt=\frac17\int\frac{\sin^2t}{\cos^2t}dt=\frac17\int\frac{1-\cos^2t}{\cos^2t}dt=\\ =\frac17\int\left\{\frac1{\cos^2t}-\frac{\cos^2t}{\cos^2t}\right\}dt=\frac17\int\left\{\frac1{\cos^2t}-1\right\}dt=\\ =\frac17\left\{\int\frac{dt}{\cos^2t}-\int dt\right\}=\frac17\left\{\tan t-t+C\right\}=\\ =\left\{t=7x\right\}=\frac17\left\{\tan7x-7x+C\right\}=\frac17\tan7x-x+C$