Tìm \(I = \int + \sin x}}\,dx} \).

* Hướng dẫn giải

Ta có:

\(I = \int {\dfrac{{{{\cos }^3}x}}{{1 + \sin x}}\,dx} \)

\(= \int {\dfrac{{{{\cos }^2}x}}{{1 + \sin x}}} \,d\left( {\sin x} \right) \)

\(= \int {\dfrac{{1 - {{\sin }^2}x}}{{1 + \sin x}}} \,d\left( {\sin x} \right)\)

\( = \int {\left( {1 - \sin x} \right)} \,d\left( {\sin x} \right) \)

\(= \left( {\sin x - \dfrac{1}{2}{{\sin }^2}x} \right) + C\)