Cho \(\int_{0}^{1} f(2 x+1) \mathrm{d} x=12 \text { và } \int_{0}^{\frac{\pi}{2}} f\left(\sin ^{2} x\right) \sin 2 x \mathrm{~d} x=3 . \text { Tính } \int_{0}^{3} f(x) \mathrm{d} x\)
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Lời giải:
Báo sai\(\begin{array}{l} \text { Đặt } 2 x+1=t \Rightarrow 12=\int_{1}^{3} f(t) \mathrm{d}\left(\frac{t-1}{2}\right)=\frac{1}{2} \int_{1}^{3} f(t) \mathrm{d} t=\frac{1}{2} \int_{1}^{3} f(x) \mathrm{d} x \Rightarrow \int_{1}^{3} f(x) \mathrm{d} x=24 \\ \text { Ta có } \int_{0}^{\frac{\pi}{2}} f\left(\sin ^{2} x\right) \sin 2 x \mathrm{~d} x=\int_{0}^{\frac{\pi}{2}} f\left(\sin ^{2} x\right) \cdot 2 \sin x \cos x \mathrm{~d} x=\int_{0}^{\frac{\pi}{2}} 2 \sin x \cdot f\left(\sin ^{2} x\right) \mathrm{d}(\sin x) \\ =\int_{0}^{\frac{\pi}{2}} f\left(\sin ^{2} x\right) \mathrm{d}\left(\sin ^{2} x\right)=\int_{0}^{1} f(u) \mathrm{d} u=\int_{0}^{1} f(x) \mathrm{d} x=3 \\ \Rightarrow \int_{0}^{3} f(x) \mathrm{d} x=\int_{0}^{1} f(x) \mathrm{d} x+\int_{1}^{3} f(x) \mathrm{d} x=3+24=27 \end{array}\)