Potrzebne pochodne, całki i tożsamości
1. $\int{\frac{1}{\text{sin}\alpha}d\alpha}=\text{ln}|\text{tg}\left(\frac{\alpha}{2}\right)|+C$
Dowód 1
$\int{\frac{1}{\text{sin}\alpha}d\alpha}=\int{\frac{1}{{2\text{sin}\frac{\alpha}{2}}\text{cos}\frac{\alpha}{2}}d\alpha}
=\int{\frac{1}{{2\text{tg}\frac{\alpha}{2}}\text{cos}^2\frac{\alpha}{2}}d\alpha} =\star
=\int{\frac{1}{t}dt}=\text{ln}|t|+C = ln|\text{tg}\left(\frac{\alpha}{2}\right)|+C
$
$\star $
$\star $
$\text{tg}\left(\frac{\alpha}{2}\right)=t $
$\frac{1}{\text{cos}^2\left(\frac{\alpha}{2}\right)}\cdot \frac{1}{2}d\alpha = dt $
$\frac{1}{\text{cos}^2\left(\frac{\alpha}{2}\right)}\cdot \frac{1}{2}d\alpha = dt $
2. $\int{\frac{1}{\text{cos}\alpha}d\alpha}=\text{ln}|\text{tg}\left(\frac{\alpha}{2}+\frac{\pi}{4}\right)|+C$
Dowód 2
$\int{\frac{1}{\text{cos}\alpha}d\alpha}=\int{\frac{1}{\text{sin}\left(\alpha+\frac{\pi}{2}\right)}d\alpha}
=\star
=\int{\frac{1}{\text{sin}t}dt} =\text{ln}|\text{tg}\left(\frac{t}{2}\right)|+C =\text{ln}|\text{tg}\left(\frac{\alpha}{2}+\frac{\pi}{4}\right)|+C
$
$\star $
$\star $
$\alpha+\frac{\pi}{2}=t $
$d\alpha = dt $
$d\alpha = dt $
3. $\int{\frac{\text{sin}^2\alpha}{\text{cos}^3\alpha}d\alpha}= \frac{\text{sin}\alpha}{2\text{cos}^2\alpha}-\frac{1}{2}\cdot \text{ln}|\text{tg}\left(\frac{\alpha}{2}+\frac{\pi}{4}\right)|+C$
Dowód 3
$\int{\frac{\text{sin}^2\alpha}{\text{cos}^3\alpha}d\alpha}=
\int{\text{sin}\alpha \cdot \frac{\text{sin}\alpha}{\text{cos}^3\alpha}d\alpha}=\star=
\text{sin}\alpha \cdot \frac{1}{2\text{cos}^2\alpha}-\int{\text{cos}\alpha \cdot \frac{1}{2\text{cos}^2\alpha}}=
\text{sin}\alpha \cdot \frac{1}{2\text{cos}^2\alpha}-\frac{1}{2}\cdot \int{\frac{1}{\text{cos}\alpha}d\alpha}=
\frac{\text{sin}\alpha}{2\text{cos}^2\alpha}-\frac{1}{2}\cdot \text{ln}|\text{tg}\left(\frac{\alpha}{2}+\frac{\pi}{4}\right)|+C
$
$\star $
$\star \star$
$\star $
$u=\text{sin}\alpha \qquad u'= \text{cos}\alpha$
$v' = \frac{\text{sin}\alpha}{\text{cos}^3\alpha} \qquad v=\int{\frac{\text{sin}\alpha}{\text{cos}^3\alpha}d\alpha} =\star \star= -\int{\frac{1}{t^3}dt}=\frac{1}{2t^2}= \frac{1}{2\text{cos}^2\alpha}$
$v' = \frac{\text{sin}\alpha}{\text{cos}^3\alpha} \qquad v=\int{\frac{\text{sin}\alpha}{\text{cos}^3\alpha}d\alpha} =\star \star= -\int{\frac{1}{t^3}dt}=\frac{1}{2t^2}= \frac{1}{2\text{cos}^2\alpha}$
$\star \star$
$\text{cos}\alpha = t$
$-\text{sin}\alpha d\alpha= dt$
$-\text{sin}\alpha d\alpha= dt$
4. $\int{\frac{1}{\text{cos}^3\alpha}d\alpha}= \frac{\text{sin}\alpha}{2\text{cos}^2\alpha}+\frac{1}{2}\cdot \text{ln}|\text{tg}\left(\frac{\alpha}{2}+\frac{\pi}{4}\right)|+C$
Dowód 4
$\int{\frac{1}{\text{cos}^3\alpha}d\alpha}=\int{\frac{\text{sin}^2\alpha+\text{cos}^2\alpha}{\text{cos}^3\alpha}d\alpha}=
\int{\frac{\text{sin}^2\alpha}{\text{cos}^3\alpha}d\alpha} + \int{\frac{1}{\text{cos}\alpha}d\alpha}=
\frac{\text{sin}\alpha}{2\text{cos}^2\alpha}-\frac{1}{2}\cdot \text{ln}|\text{tg}\left(\frac{\alpha}{2}+\frac{\pi}{4}\right)|+
\text{ln}|\text{tg}\left(\frac{\alpha}{2}+\frac{\pi}{4}\right)|+C=
\frac{\text{sin}\alpha}{2\text{cos}^2\alpha}+\frac{1}{2}\cdot \text{ln}|\text{tg}\left(\frac{\alpha}{2}+\frac{\pi}{4}\right)|+C$
5. $\text{tg}\left(\frac{\alpha}{2}+\frac{\pi}{4}\right)=\text{sec}\alpha+\text{tg}\alpha$
Dowód 5
Niech $t=\text{tg}\left(\frac{\alpha}{2}\right)$.
Wtedy $\text{sin}\left(\frac{\alpha}{2}\right)=\frac{t}{\sqrt{1+t^2}}$
oraz $\text{cos}\left(\frac{\alpha}{2}\right)=\frac{1}{\sqrt{1+t^2}}$.
Stąd:
$\text{tg}\left(\frac{\alpha}{2}+\frac{\pi}{4}\right) =\frac{\text{tg}\left(\frac{\alpha}{2}\right)+\text{tg}\left(\frac{\pi}{4}\right)}{1-\text{tg}\left(\frac{\alpha}{2}\right)\cdot \text{tg}\left(\frac{\pi}{4}\right)} =\frac{t+1}{1-t\cdot 1} = \frac{1+t}{1-t}=\frac{(1+t)\cdot (1+t)}{(1-t)\cdot (1+t)} =\frac{t^2+1+2t}{1-t^2}=\frac{\frac{t^2+1+2t}{1+t^2}}{\frac{1-t^2}{1+t^2}} =\frac{1+\text{sin}\alpha}{\text{cos}\alpha}=\text{sec}\alpha+\text{tg}\alpha$
- $\text{sin}\alpha = 2\cdot \text{sin}\left(\frac{\alpha}{2}\right)\cdot \text{cos}\left(\frac{\alpha}{2}\right) =\frac{2t}{1+t^2}$
- $\text{cos}\alpha = 2\cdot \text{cos}^2\left(\frac{\alpha}{2}\right)-\text{sin}^2\left(\frac{\alpha}{2}\right) =\frac{1-t^2}{1+t^2}$
- $\text{tg}\alpha =\frac{\text{sin}\alpha}{\text{cos}\alpha} =\frac{2t}{1-t^2}$
$\text{tg}\left(\frac{\alpha}{2}+\frac{\pi}{4}\right) =\frac{\text{tg}\left(\frac{\alpha}{2}\right)+\text{tg}\left(\frac{\pi}{4}\right)}{1-\text{tg}\left(\frac{\alpha}{2}\right)\cdot \text{tg}\left(\frac{\pi}{4}\right)} =\frac{t+1}{1-t\cdot 1} = \frac{1+t}{1-t}=\frac{(1+t)\cdot (1+t)}{(1-t)\cdot (1+t)} =\frac{t^2+1+2t}{1-t^2}=\frac{\frac{t^2+1+2t}{1+t^2}}{\frac{1-t^2}{1+t^2}} =\frac{1+\text{sin}\alpha}{\text{cos}\alpha}=\text{sec}\alpha+\text{tg}\alpha$
6. $\text{tg}\left(\frac{\alpha}{2}\right)=\frac{\text{tg}\alpha}{1+\sqrt{1+\text{tg}^2 \alpha}}$
Dowód 6
$t=\text{tg}\left(\frac{\alpha}{2}\right)$
$ t+t\cdot\sqrt{1+\text{tg}^2 \alpha}=\text{tg}\alpha $
$ t\cdot(1+\sqrt{1+\text{tg}^2 \alpha})=\text{tg}\alpha $
$t=\frac{\text{tg}\alpha}{1+\sqrt{1+\text{tg}^2 \alpha}}$
$ t+t\cdot\sqrt{1+\text{tg}^2 \alpha}=\text{tg}\alpha $
$ t\cdot(1+\sqrt{1+\text{tg}^2 \alpha})=\text{tg}\alpha $
$t=\frac{\text{tg}\alpha}{1+\sqrt{1+\text{tg}^2 \alpha}}$
7. $(\text{sec}(\alpha))'=\text{tg}(\alpha)\cdot \text{sec}(\alpha)$
Dowód 7
$(\text{sec}(\alpha))'= (\text{cos}^{-1}(\alpha))'
=-\text{sin}(\alpha)\cdot (-1)\cdot \text{cos}^{-2}(\alpha)
=\frac{\text{sin}(\alpha)}{\text{cos}^2(\alpha)}
=\text{tg}(\alpha)\cdot \text{sec}(\alpha)$
8. $\int{\cos^2(\alpha)\;d\alpha}=\frac{1}{2}\cdot(\alpha+\sin\alpha \cdot \cos \alpha)+C$
Dowód 8
$2\cos^2\alpha-1 = \cos(2\alpha)$
$2\cos^2\alpha = 1+\cos(2\alpha)$
$\cos^2\alpha =\frac{1}{2}\cdot (1+\cos(2\alpha))$
$\int{\cos^2(\alpha)\;d\alpha}=\frac{1}{2} \cdot \int{(1+\cos(2\alpha))\;d\alpha}$
$\int{\cos^2(\alpha)\;d\alpha}=\frac{1}{2}\cdot (\alpha+\frac{1}{2}\cdot \sin(2\alpha)) +C$
$\int{\cos^2(\alpha)\;d\alpha}=\frac{1}{2}\cdot (\alpha+\frac{1}{2}\cdot 2\sin \alpha\cdot \cos\alpha) +C$
$\int{\cos^2(\alpha)\;d\alpha}=\frac{1}{2}\cdot(\alpha+\sin\alpha \cdot \cos \alpha)+C$
$2\cos^2\alpha = 1+\cos(2\alpha)$
$\cos^2\alpha =\frac{1}{2}\cdot (1+\cos(2\alpha))$
$\int{\cos^2(\alpha)\;d\alpha}=\frac{1}{2} \cdot \int{(1+\cos(2\alpha))\;d\alpha}$
$\int{\cos^2(\alpha)\;d\alpha}=\frac{1}{2}\cdot (\alpha+\frac{1}{2}\cdot \sin(2\alpha)) +C$
$\int{\cos^2(\alpha)\;d\alpha}=\frac{1}{2}\cdot (\alpha+\frac{1}{2}\cdot 2\sin \alpha\cdot \cos\alpha) +C$
$\int{\cos^2(\alpha)\;d\alpha}=\frac{1}{2}\cdot(\alpha+\sin\alpha \cdot \cos \alpha)+C$
Zadanie. Zapisz równości w podpunktach 2, 3, 4 wykorzystując tylko funkcje secans i tangens oraz logarytm naturalny.
Rozwiązanie
- $\int{\text{sec}(\alpha) \text{d}\alpha}=\text{ln}|\text{sec}(\alpha)+\text{tg}(\alpha)|+C$
- $\int{\text{tg}^2(\alpha)\text{sec}(\alpha) \text{d}\alpha} =\frac{1}{2}(\text{tg}(\alpha)\cdot \text{sec}(\alpha)-\text{ln}|\text{sec}(\alpha)+\text{tg}(\alpha)|)+C$
- $\int{\text{sec}^3(\alpha) \text{d}\alpha} =\frac{1}{2}(\text{tg}(\alpha)\cdot \text{sec}(\alpha)+\text{ln}|\text{sec}(\alpha)+\text{tg}(\alpha)|)+C$