Odpowiedź:
[tex]$\frac{1}{6} -\frac{1}{8} sin(2)-\frac{1}{4} cos(2)[/tex]
Szczegółowe wyjaśnienie:
Całka:
[tex]$\int\limits^1_0 {x^{2}sin^{2}x} \, dx[/tex]
Najpierw obliczymy całkę nieoznaczoną:
[tex]$\int {x^{2}sin^{2}x} \, dx=\frac{1}{2} \int {x^{2}(1-cos2x)} \, dx =\frac{1}{2} \int {x^{2}} \, dx -\frac{1}{2}\int {x^{2}cos2x} \, dx =[/tex]
[tex]$=\frac{1}{6} x^{3}-\frac{1}{2} \int {x^{2}cos2x} \, dx =\left|\begin{array}{cc}f=x^{2}&dg=cos2x \ dx\\df=2x \ dx&g=\frac{1}{2}sin2x \end{array}\right|=[/tex]
[tex]$=\frac{1}{6}x^{3}-\frac{1}{2} \Big(\frac{1}{2}x^{2}sin2x- \int {xsin2x} \, dx \Big)=\frac{1}{6}x^{3}-\frac{1}{4}x^{2}sin2x+\frac{1}{2} \int {xsin2x} \, dx =[/tex]
[tex]$=\left|\begin{array}{cc}f=x&dg=sin2x \ dx\\df= dx&g=-\frac{1}{2}cos2x \end{array}\right|=[/tex]
[tex]$=\frac{1}{6}x^{3}-\frac{1}{4}x^{2}sin2x+\frac{1}{2} \Big(-\frac{1}{2}xcos2x +\frac{1}{2} \int cos2x\, dx \Big) =[/tex]
[tex]$=\frac{1}{6} x^{3}-\frac{1}{4} x^{2}sin2x-\frac{1}{4} xcos2x+\frac{1}{4} \int {cos2x} \, dx=[/tex]
[tex]$=\frac{1}{6} x^{3}-\frac{1}{4} x^{2}sin2x-\frac{1}{4} xcos2x+\frac{1}{8}sin2x+C[/tex]
Teraz przechodzimy do całki oznaczonej:
[tex]$\int\limits^1_0 {x^{2}sin^{2}x} \, dx=\frac{1}{6} x^{3}-\frac{1}{4} x^{2}sin2x-\frac{1}{4} xcos2x+\frac{1}{8}sin2x \Big|^{1}_{0}=[/tex]
[tex]$=\frac{1}{6} -\frac{1}{4}sin(2) -\frac{1}{4} cos(2)+\frac{1}{8}sin(2)=\frac{1}{6} -\frac{1}{8} sin(2)-\frac{1}{4} cos(2)[/tex]
