Đáp án:
$I = (x+1)^2 sin(x)|_{0->\frac{\pi}{2}} + 2(x+1)cos(x)|_{0->\frac{\pi}{2}} + 2(-sin(\frac{\pi}{2})+sin(0)) \\$
$I = arctan(x).\frac{x^2}{2}|_{0->\frac{\pi}{2}} - [\frac{x}{2}-\frac{arctan(x)}{2}]|_{0->\frac{\pi}{2}} \\$
Giải thích các bước giải: 1)
$I = \int\limits^\frac{\pi}{2}_0 {(x+1)^2cos(x)} \, dx\\
u = (x+1)^2 => du = 2(x+1)dx \\
dv = cos(x) => v = sin(x) \\ $
$I = (x+1)^2 sin(x)|_{0->\frac{\pi}{2}} - \int\limits^\frac{\pi}{2}_0 {2(x+1)sin(x)} \, dx\\$
$
a=2(x+1) => da = 2dx \\
db = sin(x)dx => b = -cos(x)
$
$I = (x+1)^2 sin(x)|_{0->\frac{\pi}{2}} + 2(x+1)cos(x)|_{0->\frac{\pi}{2}} + \int\limits^{\frac{\pi}{2}}_0 {-2cos(x)} \, dx \\$
$I = (x+1)^2 sin(x)|_{0->\frac{\pi}{2}} + 2(x+1)cos(x)|_{0->\frac{\pi}{2}} + 2(-sin(\frac{\pi}{2})+sin(0))$
2) $ I = \int\limits^{\frac{\pi}{2}}_0 {x.arctan(x)} \, dx \\$
$ u = arctan(x) => du=\frac{1}{x^2+1} dx \\
dv = x => v = \frac{x^2}{2} \\
$ $ I = (arctan(x)*\frac{x^2}{2})|_{0->\frac{\pi}{2}} - \int\limits^{\frac{\pi}{2}}_0 {\frac{x^2+1-1}{2(x^2+1)}} \, dx \\$
$ I = (arctan(x)*\frac{x^2}{2})|_{0->\frac{\pi}{2}} - \int\limits^{\frac{\pi}{2}}_0 {\frac{1}{2}-\frac{1}{2(x^2+1)}} \, dx \\$
$I = arctan(x).\frac{x^2}{2}|_{0->\frac{\pi}{2}} - [\frac{x}{2}-\frac{arctan(x)}{2}]|_{0->\frac{\pi}{2}} \\$
