a) $\displaystyle \int{ \sin{x} \cos{5x}}dx$
$\displaystyle = \int{\left[ \dfrac{1}{2} \left ( \sin{(-4x)} + \sin{6x} \right ) \right ]}dx \\
\displaystyle = \dfrac{1}{2} \left ( \int{ \sin{(-4x)}}dx + \int{ \sin{6x} }dx \right )\\
\displaystyle = \dfrac{1}{2} \int{ \sin{(-4x)}}dx + \dfrac{1}{2} \int{ \sin{6x} }dx \\
\displaystyle = \left ( - \dfrac{1}{2} \right ) \int{ \sin{4x}}dx + \dfrac{1}{2} \int{ \sin{6x} }dx \\
= \left ( - \dfrac{1}{2} \right ) \left ( - \dfrac{1}{4} \right ) \cos{4x} + \dfrac{1}{2} \left ( - \dfrac{1}{6} \right ) \cos{6x} +C \\
= \dfrac{1}{8} \cos{4x} - \dfrac{1}{12} \cos{6x}+C$
(hoặc $ = \dfrac{1}{24}(3 \cos{4x}-2 \cos{6x}) + C$)
b) $\displaystyle \int{ \dfrac{ \tan{x}}{ \cos^2{x}}}dx$
Đặt $t= \tan{x} \Rightarrow dt = \dfrac{dx}{\cos^2{x}}$
Do đó ta có: $\displaystyle \int{t}dt = \dfrac{t^2}{2} + C =\dfrac{ \tan^2{x}}{2}+C$
Vậy $\displaystyle \int{ \dfrac{ \tan{x}}{ \cos^2{x}}}dx = \dfrac{ \tan^2{x}}{2}+C$
e) $\displaystyle \int {\sin{\left ( 2x - \dfrac{ \pi}{3} \right )}}dx$
$= \displaystyle \int {\left ( \sin{2x} \cos{ \dfrac{ \pi}{3} } - \cos{2x} \sin{ \dfrac{ \pi}{3} } \right ) }dx \\
= \displaystyle \int {\left ( \dfrac{1}{2} \sin{2x} - \dfrac{ \sqrt{3}}{2} \cos{2x} \right ) }dx \\
= \dfrac{1}{2} \displaystyle \int { \sin{2x}}dx - \displaystyle \dfrac{ \sqrt{3}}{2} \int { \cos{2x}}dx \\
= \dfrac{1}{2} \left ( - \dfrac{1}{2} \right ) \cos{2x} - \dfrac{ \sqrt{3}}{2} . \dfrac{1}{2} \sin{2x} +C \\
= - \dfrac{1}{4} \cos{2x} - \dfrac{ \sqrt{3}}{4} \sin{2x} +C$
(hoặc $ = - \dfrac{1}{2} \sin{ \left ( 2x+ \dfrac{ \pi}{6} \right ) }+ C$)
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
