a) Đặt \(u = \sqrt {7 - 3{x^2}} \Rightarrow {u^2} = 7 - 3{x^2} \Rightarrow 2udu = - 6xdx \Rightarrow 3xdx = - udu\)
Do đó \(\int {3x\sqrt {7 - 3{x^2}} dx = - \int {{u^2}du = - {{{u^3}} \over 3} + C} = - {1 \over 3}\sqrt {{{\left( {7 - 3{x^2}} \right)}^3}} + C} \)
b) \(\int {\cos \left( {3x + 4} \right)dx = {1 \over 3}\sin \left( {3x + 4} \right) + C} \)
c) \(\int {{{dx} \over {{{\cos }^2}\left( {3x + 2} \right)}} = {1 \over 3}\tan \left( {3x + 2} \right) + C} \)
d) Đặt \(u = \sin {x \over 3} \Rightarrow du = {1 \over 3}\cos {x \over 3}dx \Rightarrow \cos {x \over 3}dx = 3du\)
Do đó \(\int {{{\sin }^5}{x \over 3}\cos {x \over 3}dx = 3\int {{u^5}du = {{{u^6}} \over 2} + C = {1 \over 2}{{\sin }^6}\left( {{x \over 3}} \right) + C.} } \)
