całka nie do rozwiązania

[siebka]

męczę sie i nie moje rozwiązać tej całki

\(\displaystyle{ \int\frac{1}{\sin(x)+7\cos(x)}dx}\)

[luka52]

Podstawmy \(\displaystyle{ t = \tan \frac{x}{2}, \quad dx = \frac{2dt}{1+t^2}}\)
\(\displaystyle{ \sin x = \frac{2t}{1+t^2}, \quad \cos x = \frac{1-t^2}{1+t^2}}\)
\(\displaystyle{ I = t \frac{1}{\frac{2t}{1+t^2} + 7 \frac{1-t^2}{1+t^2}} \frac{2dt}{1+t^2} = -2 t \frac{dt}{7t^2 - 2t - 7} = \ldots}\)

[max]

Albo takowe, nieco cudaczne przekształcenie:
\(\displaystyle{ \sin x + 7\cos x = \sqrt{1^{2} + 7^{2}}\left(\tfrac{1}{\sqrt{1^{2} + 7^{2}}}\sin x + \tfrac{7}{\sqrt{1^{2} + 7^{2}}}\cos x\right) =\\
= \sqrt{50}(\cos \varphi \sin x + \sin \varphi \cos x) = \sqrt{50}\sin(x + \varphi)}\)
stąd:
\(\displaystyle{ \int \frac{dx}{\sin x + 7\cos x} = \frac{1}{\sqrt{50}}\int \frac{dx}{\sin (x + \varphi)} =\\ =\frac{1}{\sqrt{50}}\ln \left|\tan \left(\frac{x + \varphi}{2}\right)\right| + C = \frac{1}{\sqrt{50}}\ln \left|\frac{1 - \cos (x + \varphi)}{\sin (x + \varphi)}\right| + C = \\ =\frac{1}{\sqrt{50}}\ln \left|\frac{1 - \frac{1}{\sqrt{50}}\cos x + \frac{7}{\sqrt{50}}\sin x}{\frac{1}{\sqrt{50}}\sin x + \frac{7}{\sqrt{50}}\cos x}\right| + C =\\
=\frac{1}{\sqrt{50}}\ln \left|\frac{\sqrt{50} - \cos x + 7\sin x}{\sin x + 7\cos x}\right| + C}\)