Całka nieoznaczona
: 6 maja 2008, o 17:04
jak policzyc taka calke?
\(\displaystyle{ \int_{}^{} xsin^2x dx}\)
\(\displaystyle{ \int_{}^{} xsin^2x dx}\)
\(\displaystyle{ \sin^2x=\frac{1-\cos 2x}{2}}\)
\(\displaystyle{ \int \sin^2 x \ dx = t \frac{1-\cos 2x}{2} dx = \frac{1}{2} t dx - \frac{1}{2} t \cos 2x \ dx = \frac{x}{2} - \frac{\sin 2x}{4} + C}\)
\(\displaystyle{ \int x \sin^2x \ dx =... \\
u=x \qquad dv=\sin^2 x \ dx \\
du=dx \qquad v = \frac{x}{2} - \frac{\sin 2x}{4} \\
... = \frac{1}{2}x^2 - \frac{1}{4}x\sin 2x - t ft( \frac{x}{2} - \frac{\sin 2x}{4} \right)dx = \\
= \frac{1}{2}x^2 - \frac{1}{4}x\sin 2x - \frac{1}{4}x^2 - \frac{1}{8}\cos 2x + C = \\
= \frac{1}{4}x^2 - \frac{1}{8}\cos 2x - \frac{1}{4}x\sin 2x + C}\)