Bài tập 19 trang 161 SGK Toán 12 NC

Bài tập 19 trang 161 SGK Toán 12 NC

a) Đặt \(u = \sqrt {{t^5} + 2t}  \Rightarrow {u^2} = {t^5} + 2t \)

\(\Rightarrow 2udu = (5{t^4} + 2)dt\)

\(\begin{array}{l}
\int\limits_0^1 {\sqrt {{t^5} + 2t} } \left( {2 + 5{t^4}} \right)dt\\
 = \int\limits_0^{\sqrt 3 } 2 {u^2}du = \left. {\frac{{2{u^3}}}{3}} \right|_0^{\sqrt 3 } = 2\sqrt 3 
\end{array}\)

b) Ta có:

\(\int\limits_0^{\frac{\pi }{2}} x \sin x\cos xdx = \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} x \sin 2xdx\)

Đặt \(\left\{ \begin{array}{l}
u = x\\
dv = sin2xdx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = dx\\
v =  - \frac{1}{2}\cos 2x
\end{array} \right.\)

Do đó:

\(\begin{array}{l}
I = \left. {\frac{1}{2}\left( { - \frac{1}{2}\cos 2x} \right)} \right|_0^{\frac{\pi }{2}} + \frac{1}{4}\int\limits_0^{\frac{\pi }{2}} {\cos 2x} dx\\
 = \frac{\pi }{8} + \frac{1}{8}\left. {\sin 2x} \right|_0^{\frac{\pi }{2}} = \frac{\pi }{8}
\end{array}\)

-- Mod Toán 12