Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
I = \int\limits_0^{\frac{\pi }{2}} {\frac{1}{{\sin x}}dx} = \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin x}}{{{{\sin }^2}x}}dx} = \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin x}}{{1 - {{\cos }^2}x}}dx} = \int\limits_0^{\frac{\pi }{2}} {\frac{{ - \left( {\sin xdx} \right)}}{{{{\cos }^2}x - 1}}} \\
t = \cos x \Rightarrow \left\{ \begin{array}{l}
dt = \left( {\cos x} \right)'dx = - \sin xdx\\
x = 0 \Rightarrow t = 1\\
x = \frac{\pi }{2} \Rightarrow t = 0
\end{array} \right.\\
\Rightarrow I = \int\limits_1^0 {\frac{{dt}}{{{t^2} - 1}}} \\
= - \int\limits_0^1 {\frac{1}{{\left( {t - 1} \right)\left( {t + 1} \right)}}dt} \\
= - \frac{1}{2}.\int\limits_0^1 {\frac{2}{{\left( {t - 1} \right)\left( {t + 1} \right)}}dt} \\
= - \frac{1}{2}.\int\limits_0^1 {\frac{{\left( {t + 1} \right) - \left( {t - 1} \right)}}{{\left( {t - 1} \right)\left( {t + 1} \right)}}dt} \\
= - \frac{1}{2}.\int\limits_0^1 {\left[ {\frac{1}{{t - 1}} - \frac{1}{{t + 1}}} \right]dt} \\
= - \frac{1}{2}.\mathop {\left. {\left( {\ln \left| {t - 1} \right| - \ln \left| {t + 1} \right|} \right)} \right|}\nolimits_0^1 \\
= - \frac{1}{2}.\left( {\ln 0 - \ln 1 - \ln 2 + \ln 1} \right)
\end{array}\)
(Không tồn tại \(\ln 0\) )
