Tính \(\displaystyle \int {\frac{1}{{\sin x - \sin a}}} dx\)
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Lời giải:
Báo saiTa có: \(\displaystyle \frac{1}{{\sin x - \sin a}}\)\(\displaystyle = \frac{1}{{2\cos \frac{{x + a}}{2}\sin \frac{{x - a}}{2}}}\) \(\displaystyle = \frac{{\cos a}}{{2\cos a\cos \frac{{x + a}}{2}\sin \frac{{x - a}}{2}}}\)
\(\displaystyle = \frac{{\cos \left( {\frac{{x + a}}{2} - \frac{{x - a}}{2}} \right)}}{{2\cos a\cos \frac{{x + a}}{2}\sin \frac{{x - a}}{2}}}\) \(\displaystyle = \frac{{\cos \frac{{x + a}}{2}\cos \frac{{x - a}}{2} + \sin \frac{{x + a}}{2}\sin \frac{{x - a}}{2}}}{{2\cos a\cos \frac{{x + a}}{2}\sin \frac{{x - a}}{2}}}\)
\(\displaystyle = \frac{1}{{2\cos a}}\left( {\frac{{\cos \frac{{x - a}}{2}}}{{\sin \frac{{x - a}}{2}}} + \frac{{\sin \frac{{x + a}}{2}}}{{\cos \frac{{x + a}}{2}}}} \right)\)
\(\displaystyle \Rightarrow \int {\frac{1}{{\sin x - \sin a}}} dx\) \(\displaystyle = \frac{1}{{2\cos a}}\int {\left( {\frac{{\cos \frac{{x - a}}{2}}}{{\sin \frac{{x - a}}{2}}} + \frac{{\sin \frac{{x + a}}{2}}}{{\cos \frac{{x + a}}{2}}}} \right)dx} \)
+) Tính \(\displaystyle J = \int {\frac{{\cos \frac{{x - a}}{2}}}{{\sin \frac{{x - a}}{2}}}dx} \) \(\displaystyle = \int {\frac{{2d\left( {\sin \frac{{x - a}}{2}} \right)}}{{\sin \frac{{x - a}}{2}}}} \) \(\displaystyle = 2\ln \left| {\sin \frac{{x - a}}{2}} \right| + D\)
+) Tính \(\displaystyle K = \int {\frac{{\sin \frac{{x + a}}{2}}}{{\cos \frac{{x + a}}{2}}}dx} \) \(\displaystyle = \int {\frac{{ - 2d\left( {\cos \frac{{x + a}}{2}} \right)}}{{\cos \frac{{x + a}}{2}}}} \) \(\displaystyle = - 2\ln \left| {\cos \frac{{x + a}}{2}} \right| + D\)
\(\displaystyle \Rightarrow I = \frac{1}{{2\cos a}}\left( {J + K} \right)\) \(\displaystyle = \frac{1}{{2\cos a}}\left( {2\ln \left| {\sin \frac{{x - a}}{2}} \right| - 2\ln \left| {\cos \frac{{x + a}}{2}} \right|} \right) + C\) \(\displaystyle = \frac{1}{{\cos a}}\ln \left| {\frac{{\sin \frac{{x - a}}{2}}}{{\cos \frac{{x + a}}{2}}}} \right| + C\)
