Tìm \(I = \int {\left( {x - 2\sin x} \right)\frac{{dx}}{{{\rm{co}}{{\rm{s}}^2}x}}} \)

* Hướng dẫn giải

Ta có \(\begin{array}{l}
I = \int {\left( {x - 2\sin x} \right)\frac{{dx}}{{{\rm{co}}{{\rm{s}}^2}x}}}  = \int {x\frac{{dx}}{{{\rm{co}}{{\rm{s}}^2}x}}}  - \int {2\sin x\frac{1}{{{\rm{co}}{{\rm{s}}^2}x}}} dx= {I_1} - {I_2}
\end{array}\)

Xét \({I_1} = \int {x\frac{{dx}}{{{\rm{co}}{{\rm{s}}^2}x}}} \)

Đặt \(\left\{ \begin{array}{l}
u = x\\
dv = \frac{{dx}}{{{{\cos }^2}x}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = dx\\
v = \tan x
\end{array} \right.\)

\(\begin{array}{l}
{I_1} = x\tan x - \int {\tan xdx = x\tan x - \int {\frac{{\sin x}}{{\cos x}}} dx} \\
 = x\tan x - \int {\frac{{d\left( {\sin x} \right)}}{{\cos x}} = } x\tan x + \ln \cos x + C
\end{array}\)

\){I_2} = \int {2\sin x\frac{1}{{{{\cos }^2}x}}} dx = \int {\frac{{ - 2d\left( {\cos x} \right)}}{{{{\cos }^2}x}}}  = \frac{2}{{\cos x}} + C\)