Ściąga - co trzeba wiedzieć
- $(\text{sec}(\alpha))'=\text{tg}(\alpha)\cdot \text{sec}(\alpha)$
- $\int{\text{cos}^2(\alpha) \;d\alpha}=\frac{1}{2}\cdot (\alpha+\sin\alpha\cdot cos\alpha)+C$
- $\int{\text{sec}(\alpha) \;d\alpha}=\text{ln}|\text{sec}(\alpha)+\text{tg}(\alpha)|+C$
- $\int{\text{tg}^2(\alpha)\text{sec}(\alpha) \;d\alpha} =\frac{1}{2}(\text{tg}(\alpha)\cdot \text{sec}(\alpha)-\text{ln}|\text{sec}(\alpha)+\text{tg}(\alpha)|)+C$
- $\int{\text{sec}^3(\alpha) \;d\alpha} =\frac{1}{2}(\text{tg}(\alpha)\cdot \text{sec}(\alpha)+\text{ln}|\text{sec}(\alpha)+\text{tg}(\alpha)|)+C$
A. Obliczanie całek postaci $\int{\sqrt{x^2-a^2}\;dx}$ oraz $\int{\frac{1}{\sqrt{x^2-a^2}}\;dx}$
$x = a\cdot \sec(\alpha)$.
$\sqrt{x^2 - a^2} = a\cdot \text{tg}(\alpha)$.
$dx = a\cdot \text{tg}(\alpha)\cdot \text{sec}(\alpha)\;d\alpha $
$\sqrt{x^2 - a^2} = a\cdot \text{tg}(\alpha)$.
$dx = a\cdot \text{tg}(\alpha)\cdot \text{sec}(\alpha)\;d\alpha $
$\int{\sqrt{x^2-a^2}\;dx} = a^2\cdot\int{\text{tg}^2(\alpha)\text{sec}(\alpha) \;d\alpha}$
$\int{\sqrt{x^2-a^2}\;dx} = a^2\cdot\int{\text{tg}^2(\alpha)\text{sec}(\alpha) \;d\alpha}$
$\int{\sqrt{x^2-a^2}\;dx} = a^2\cdot \frac{1}{2}(\frac{\sqrt{x^2-a^2}}{a}\cdot \frac{x}{a}-\text{ln}|\frac{x}{a}+\frac{\sqrt{x^2-a^2}}{a}|)+C$
$\int{\sqrt{x^2-a^2}\;dx} = a^2\cdot \frac{1}{2}(\frac{x\sqrt{x^2-a^2}}{a^2}-\text{ln}|\frac{x+\sqrt{x^2-a^2}}{a}|)+C$
$\int{\sqrt{x^2-a^2}\;dx} = a^2\cdot\int{\text{tg}^2(\alpha)\text{sec}(\alpha) \;d\alpha}$
$\int{\sqrt{x^2-a^2}\;dx} = a^2\cdot \frac{1}{2}(\frac{\sqrt{x^2-a^2}}{a}\cdot \frac{x}{a}-\text{ln}|\frac{x}{a}+\frac{\sqrt{x^2-a^2}}{a}|)+C$
$\int{\sqrt{x^2-a^2}\;dx} = a^2\cdot \frac{1}{2}(\frac{x\sqrt{x^2-a^2}}{a^2}-\text{ln}|\frac{x+\sqrt{x^2-a^2}}{a}|)+C$
$\int{\sqrt{x^2-a^2}\;dx} = \frac{1}{2}(x\sqrt{x^2-a^2}-a^2\cdot\text{ln}|x+\sqrt{x^2-a^2}|)+C'$
$\int{\frac{1}{\sqrt{x^2-a^2}}\;dx} = \int{\frac{1}{a\cdot \text{tg}\alpha}\cdot a\cdot \text{tg}(\alpha)\cdot \text{sec}(\alpha)\;d\alpha}$
$\int{\frac{1}{\sqrt{x^2-a^2}}\;dx} = \int{\text{sec}(\alpha)\;d\alpha}$
$\int{\frac{1}{\sqrt{x^2-a^2}}\;dx} = \text{ln}|\text{sec}(\alpha)+\text{tg}(\alpha)|+C$
$\int{\frac{1}{\sqrt{x^2-a^2}}\;dx} = \text{ln}|\frac{x+\sqrt{x^2-a^2}}{a}|+C$
$\int{\frac{1}{\sqrt{x^2-a^2}}\;dx} = \int{\text{sec}(\alpha)\;d\alpha}$
$\int{\frac{1}{\sqrt{x^2-a^2}}\;dx} = \text{ln}|\text{sec}(\alpha)+\text{tg}(\alpha)|+C$
$\int{\frac{1}{\sqrt{x^2-a^2}}\;dx} = \text{ln}|\frac{x+\sqrt{x^2-a^2}}{a}|+C$
$\int{\frac{1}{\sqrt{x^2-a^2}}\;dx} = \text{ln}|x+\sqrt{x^2-a^2}|+C'$
B. Obliczanie całek postaci $\int{\sqrt{a^2-x^2}\;dx}$ oraz $\int{\frac{1}{\sqrt{a^2-x^2}}\;dx}$
$x = a\cdot \sin(\alpha)$.
$\sqrt{a^2 - x^2} = a\cdot \text{cos}(\alpha)$.
$dx = a\cdot \cos(\alpha)\;d\alpha $
$\sqrt{a^2 - x^2} = a\cdot \text{cos}(\alpha)$.
$dx = a\cdot \cos(\alpha)\;d\alpha $
$\int{\sqrt{a^2-x^2}\;dx} = a^2\cdot\int{\text{cos}^2(\alpha)\;d\alpha}$
$\int{\sqrt{a^2-x^2}\;dx} = a^2\cdot \frac{1}{2}\cdot (\alpha +\sin\alpha \cdot \cos \alpha)+C$
$\int{\sqrt{a^2-x^2}\;dx} = a^2\cdot \frac{1}{2}\cdot \left(\arcsin(\frac{x}{a}) +\frac{x}{a}\cdot\frac{\sqrt{a^2-x^2}}{a}\right)+C$
$\int{\sqrt{a^2-x^2}\;dx} = a^2\cdot \frac{1}{2}\cdot (\alpha +\sin\alpha \cdot \cos \alpha)+C$
$\int{\sqrt{a^2-x^2}\;dx} = a^2\cdot \frac{1}{2}\cdot \left(\arcsin(\frac{x}{a}) +\frac{x}{a}\cdot\frac{\sqrt{a^2-x^2}}{a}\right)+C$
$\int{\sqrt{a^2-x^2}\;dx} = \frac{1}{2}\cdot \left(a^2\cdot \arcsin(\frac{x}{a}) +x\cdot \sqrt{a^2-x^2}\right)+C$
$\int{\frac{1}{\sqrt{a^2-x^2}}\;dx} = \int{\frac{1}{a}\cdot\text{sec}\alpha \cdot a \cdot \cos(\alpha)\;d\alpha}$
$\int{\frac{1}{\sqrt{a^2-x^2}}\;dx} = \int{d\alpha}$
$\int{\frac{1}{\sqrt{a^2-x^2}}\;dx} = \alpha + C$
$\int{\frac{1}{\sqrt{a^2-x^2}}\;dx} = \int{d\alpha}$
$\int{\frac{1}{\sqrt{a^2-x^2}}\;dx} = \alpha + C$
$\int{\frac{1}{\sqrt{x^2-a^2}}\;dx} = \arcsin \left(\frac{x}{a}\right)+C$
C. Obliczanie całek postaci $\int{\sqrt{x^2+a^2}\;dx}$ oraz $\int{\frac{1}{\sqrt{x^2+a^2}}\;dx}$
$x = a\cdot \text{tg}(\alpha)$.
$\sqrt{x^2 + a^2} = a\cdot \sec(\alpha)$.
$dx = a\cdot \sec^2(\alpha)\;d\alpha $
$\sqrt{x^2 + a^2} = a\cdot \sec(\alpha)$.
$dx = a\cdot \sec^2(\alpha)\;d\alpha $
$\int{\sqrt{x^2+a^2}\;dx} = a^2\cdot\int{\text{sec}^3(\alpha)\;d\alpha}$
$\int{\sqrt{x^2+a^2}\;dx} =a^2\cdot \left(\frac{1}{2}(\text{tg}(\alpha)\cdot \text{sec}(\alpha)+\text{ln}|\text{sec}(\alpha)+\text{tg}(\alpha)|)\right)+C $
$\int{\sqrt{x^2+a^2}\;dx} =a^2\cdot \left(\frac{1}{2}(\frac{x}{a}\cdot \frac{\sqrt{x^2+a^2}}{a}+ \text{ln}|\frac{\sqrt{x^2+a^2}}{a}+\frac{x}{a}|)\right)+C $
$\int{\sqrt{x^2+a^2}\;dx} =a^2\cdot \left(\frac{1}{2}( \frac{x\sqrt{x^2+a^2}}{a^2}+ \text{ln}|\frac{x+\sqrt{x^2+a^2}}{a}|)\right)+C $
$\int{\sqrt{x^2+a^2}\;dx} =a^2\cdot \left(\frac{1}{2}(\text{tg}(\alpha)\cdot \text{sec}(\alpha)+\text{ln}|\text{sec}(\alpha)+\text{tg}(\alpha)|)\right)+C $
$\int{\sqrt{x^2+a^2}\;dx} =a^2\cdot \left(\frac{1}{2}(\frac{x}{a}\cdot \frac{\sqrt{x^2+a^2}}{a}+ \text{ln}|\frac{\sqrt{x^2+a^2}}{a}+\frac{x}{a}|)\right)+C $
$\int{\sqrt{x^2+a^2}\;dx} =a^2\cdot \left(\frac{1}{2}( \frac{x\sqrt{x^2+a^2}}{a^2}+ \text{ln}|\frac{x+\sqrt{x^2+a^2}}{a}|)\right)+C $
$\int{\sqrt{x^2+a^2}\;dx} = \frac{1}{2}\left(x\sqrt{x^2+a^2}+a^2\cdot
\text{ln}|x+\sqrt{x^2+a^2}|\right)+C' $
$\int{\frac{1}{\sqrt{x^2+a^2}}\;dx} = \int{\frac{1}{a\cdot \sec\alpha}\cdot a\cdot \sec^2(\alpha)\;d\alpha}$
$\int{\frac{1}{\sqrt{x^2+a^2}}\;dx} = \int{\text{sec}(\alpha)\;d\alpha}$
$\int{\frac{1}{\sqrt{x^2+a^2}}\;dx} = \text{ln}|\text{sec}(\alpha)+\text{tg}(\alpha)|+C$
$\int{\frac{1}{\sqrt{x^2+a^2}}\;dx} = \text{ln}|\frac{x+\sqrt{x^2+a^2}}{a}|+C$
$\int{\frac{1}{\sqrt{x^2+a^2}}\;dx} = \int{\text{sec}(\alpha)\;d\alpha}$
$\int{\frac{1}{\sqrt{x^2+a^2}}\;dx} = \text{ln}|\text{sec}(\alpha)+\text{tg}(\alpha)|+C$
$\int{\frac{1}{\sqrt{x^2+a^2}}\;dx} = \text{ln}|\frac{x+\sqrt{x^2+a^2}}{a}|+C$
$\int{\frac{1}{\sqrt{x^2+a^2}}\;dx} = \text{ln}|x+\sqrt{x^2+a^2}|+C'$