Tìm nguyên hàm của các hàm số sau:
\(\begin{array}{*{20}{l}}
{a)f\left( x \right) = 3x\sqrt {7 - 3{x^2}} }\\
{b)f\left( x \right) = \cos \left( {3x + 4} \right)}\\
{c)f\left( x \right) = - \frac{1}{{{{\cos }^2}\left( {3x + 2} \right)}}}\\
{d)f(x) = {{\sin }^5}\frac{x}{3}\cos \frac{x}{3}}
\end{array}\)
a) Đặt \(u = \sqrt {7 - 3{x^2}} \Rightarrow {u^2} = 7 - 3{x^2} \)
\(\Rightarrow 2udu = - 6xdx \Rightarrow 3xdx = - udu\)
Do đó:
\(\begin{array}{l}
\int 3 x\sqrt {7 - 3{x^2}} dx = - \int {{u^2}} du\\
= - \frac{{{u^3}}}{3} + C = - \frac{1}{3}\sqrt {{{\left( {7 - 3{x^2}} \right)}^3}} + C
\end{array}\)
b)
\(\int \cos \left( {3x + 4} \right)dx = \frac{1}{3}\sin \left( {3x + 4} \right) + C\)
c)
\(\int \frac{{dx}}{{{{\cos }^2}\left( {3x + 2} \right)}} = \frac{1}{3}\tan \left( {3x + 2} \right) + C\)
d) Đặt \(u = \sin \frac{x}{3} \Rightarrow du = \frac{1}{3}\cos \frac{x}{3}dx \)
\(\Rightarrow \cos \frac{x}{3}dx = 3du\)
Do đó:
\(\begin{array}{l}
\int s i{n^5}\frac{x}{3}\cos \frac{x}{3}dx\\
= 3\int {{u^5}} du = \frac{{{u^6}}}{2} + C\\
= \frac{1}{2}{\sin ^6}\left( {\frac{x}{3}} \right) + C.
\end{array}\)
-- Mod Toán 12
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