Ngân hàng bài tập
A
Biết $\displaystyle\displaystyle\int\limits f(t)\mathrm{\,d}t=t^2+3t+C$. Tính $\displaystyle\displaystyle\int\limits f\left(\sin2x\right)\cos2x\mathrm{\,d}x$.
 | $\displaystyle\displaystyle\int f\left(\sin2x\right)\cos2x\mathrm{\,d}x=2\sin^2x+6\sin{x}+C$ |
 | $\displaystyle\displaystyle\int f\left(\sin2x\right)\cos2x\mathrm{\,d}x=2\sin^22x+6\sin2x+C$ |
 | $\displaystyle\displaystyle\int f\left(\sin2x\right)\cos2x\mathrm{\,d}x=\dfrac{1}{2}\sin^22x+\dfrac{3}{2}\sin2x+C$ |
 | $\displaystyle\displaystyle\int f\left(\sin2x\right)\cos2x\mathrm{\,d}x=\sin^22x+3\sin2x+C$ |
1 lời giải
Chọn phương án C.
Đặt $t=\sin2x\Rightarrow\mathrm{\,d}t=2\cos2x\mathrm{\,d}x$.
Khi đó $$\begin{aligned}\displaystyle\int\limits f(\sin2x)\cos2x\mathrm{\,d}x&=\dfrac{1}{2}\displaystyle\int\limits f(t)\mathrm{\,d}t\\ &=\dfrac{1}{2}t^2+\dfrac{3}{2}t+C\\ &=\dfrac{1}{2}\sin^22x+\dfrac{3}{2}\sin2x+C.\end{aligned}$$