Tính tích phân sau: I=∫(2x+1)cos2xdx.

Question

Tính tích phân sau:

\(I=\int _{0}^{1}\left(2x+1\right)\cos ^{2} x{\rm d}x . \)

Answer 1

\(\int _{0}^{1}\left(2x+1\right)\cos ^{2} xdx .  \)
\(\int _{0}^{1}\left(2x+1\right)\cos ^{2} xdx =\frac{1}{2} \int _{0}^{1}\left(2x+1\right)\left(1+\cos 2x\right)dx \)

\(=\frac{1}{2} \int _{0}^{1}\left(2x+1\right)dx +\frac{1}{2} \int _{0}^{1}\left(2x+1\right)\cos 2xdx   \)
\(\frac{1}{2} \int _{0}^{1}\left(2x+1\right)dx =1\)

\(I=\int _{0}^{1}\left(2x+1\right)\cos 2xdx \)

\( Đặt \left\{\begin{array}{l} {u=2x+1} \\ {dv=\cos 2xdx} \end{array}\right.\)

\(\Rightarrow \left\{\begin{array}{l} {du=2dx} \\ {v=\frac{1}{2} \sin 2x} \end{array}\right. \)
\(I=\frac{1}{2} \left(2x+1\right)\sin 2x\left|\begin{array}{l} {1} \\ {0} \end{array}\right. -\int _{0}^{1}\sin 2xdx  \)

\(=\frac{3}{2} \sin 2-\frac{1}{2} \cos 2x\left|\begin{array}{l} {1} \\ {0} \end{array}\right. =\frac{3}{2} \sin 2-\frac{1}{2} \cos 2+\frac{1}{2}  \)
Vậy: \(\int _{0}^{1}\left(2x+1\right)\cos ^{2} xdx =\frac{3}{2} \sin 2-\frac{1}{2} \cos 2+\frac{3}{2} \)