Question

Найти неопределенный интеграл x*tg^2 x dx

Answer 1

$\int\limits {x*tg^2(x)} \, dx = \int\limits {x* sin^2(x)*\frac{1}{cos^2(x)} } \, dx = \int\limits {x*sin^2(x)} \, d[tg(x)] =\\ =x*sin^2(x)*tg(x)- \int\limits {tg(x)} \, d[x*sin^2(x)] =\\ =x*sin^2(x)*tg(x)- \\ -\int\limits {tg(x)}[sin^2(x)+x*2sin(x)cos(x)] \, dx=\\ =x*sin^2(x)*tg(x)- \\ -\int\limits {tg(x)}sin^2(x) \,dx-2\int\limits {x*tg(x)*sin(x)cos(x) \, dx=\\$

$=x*sin^2(x)*tg(x)-\int\limits {\frac{sin^3(x)}{cos(x)}} \,dx-2\int\limits {x*sin^2(x) \, dx=\\$

$=x*sin^2(x)*tg(x)+\int\limits {\frac{sin^2(x)}{cos(x)}} \,d[cos(x)]-2\int\limits {x* \frac{1-cos(2x)}{2} \, dx=\\$

$=x*sin^2(x)*tg(x)+\int\limits {\frac{1-cos^2(x)}{cos(x)}} \,d[cos(x)]-[ \int\limits {x} \, dx - \int\limits {xcos(2x)} \, dx ]=$

$=x*sin^2(x)*tg(x)+\int\limits {(\frac{1}{t}-t)} \,dt-\\-[ \frac{x^2}{2} - \frac{1}{4} \int\limits {2x} \, d[sin(2x)] ]=$

$=x*sin^2(x)*tg(x)+ln|t|- \frac{t^2}{2} -\\-[ \frac{x^2}{2} - \frac{1}{4} (2x*sin(2x)- \int\limits {sin(2x)} \, d[2x] )]=$

$=x*sin^2(x)*tg(x)+ln|cos(x)|- \frac{cos^2(x)}{2} -\\-[ \frac{x^2}{2} - \frac{1}{4} (2x*sin(2x)+cos(2x) )]+C$