całka secx

[tr0jan]

Musze obliczyc całke \(\displaystyle{ \int \frac{dx}{\cos x}}\) robilem to tak (podpowiedziano mi): \(\displaystyle{ \frac{1}{cosx}*\frac{1+sinx}{1+sinx}}\)=\(\displaystyle{ \frac{secx(secx+tanx)}{secx(1+sinx)}}\)
ale nie mam pojecia co dalej!!!

Zamiast

Kod: Zaznacz cały

[tex]\int[/tex][tex]\frac{1}{cosx}[/tex]

pisz

Kod: Zaznacz cały

[tex]\int \frac{1}{cosx}[/tex]

luka52

[luka52]

IMHO lepiej podstawić \(\displaystyle{ t = \tan \frac{x}{2}, \quad dx = \frac{2 \, dt}{1+t^2}, \quad \cos x = \frac{1-t^2}{1+t^2}}\)
Co sprowadzi całkę do postaci:
\(\displaystyle{ \int \frac{1+t^2}{1-t^2} \frac{2 \, dt}{1+t^2} = 2 \int \frac{dt}{1-t^2} = \ln \left| \frac{t+1}{t-1} \right| + C = \ln \left| \frac{\tan \frac{x}{2}+1}{\tan \frac{x}{2}-1} \right| + C}\)

[max]

Można też nieco inaczej:
\(\displaystyle{ \int \frac{\mbox{d}x}{\cos x} = \int \frac{\mbox{d}x}{\sin(\frac{\pi}{2} + x)} = \int \frac{\mbox{d}x}{2\sin(\frac{\pi}{4} + \frac{x}{2})\cos(\frac{\pi}{4} + \frac{x}{2})} =\\
=\int \frac{\mbox{d}x}{2\cos^{2}(\frac{\pi}{4} + \frac{x}{2})} \frac{1}{\tan (\frac{\pi}{4} + \frac{x}{2})} \stackrel{t = \tan (\frac{\pi}{4} + \frac{x}{2})}{=} \int \frac{\mbox{d}t}{t} = \\ =\ln |t| + C = \ln \left|\tan (\tfrac{\pi}{4} + \tfrac{x}{2})\right| + C }\)

[tr0jan]

a mozesz mi jeszcze tylko wytlumaczyc skad sie wzielo \(\displaystyle{ dx=\frac{2dt}{1+t^{2}}}\)

[luka52]

\(\displaystyle{ t = \tan \frac{x}{2}\\ \frac{x}{2} = \arctan t \\ \frac{dx}{dt} = \frac{2}{1+t^2} \\ dx = \frac{2 \, dt}{1+t^2}}\)