Prośba o sprawdzenie rozwiązania
: 9 wrz 2007, o 10:35
za \(\displaystyle{ \sqrt{x} = t}\) potem u(t)= t u'(t)=1 v'(x)= cos(x) v'(x)= sin(x)
\(\displaystyle{ \int\frac{\cos\sqrt{x}}{\sqrt{x}}dx = 2\int\frac{\cos t}{t}dt =2\cdot [t\cdot \cos t -\int\sin t dt] = 2 \cdot\cos \sqrt{x}\cdot (\sqrt{x} + 1) +c}\)
\(\displaystyle{ \int\frac{\cos\sqrt{x}}{\sqrt{x}}dx = 2\int\frac{\cos t}{t}dt =2\cdot [t\cdot \cos t -\int\sin t dt] = 2 \cdot\cos \sqrt{x}\cdot (\sqrt{x} + 1) +c}\)