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Câu hỏi: Cho hàm số $f(x)$ liên tục trên $\mathbb{R}$ và thỏa mãn $\int_{\dfrac{\pi}{4}}^{\dfrac{\pi}{2}} \cot x . f\left(\sin ^2 x\right) \mathrm{d} x=\int_1^{16} \dfrac{f(\sqrt{x})}{x} \mathrm{~d} x=1$. Tính tích phân $\int_{\dfrac{1}{8}}^1 \dfrac{f(4 x)}{x} \mathrm{~d} x$.
A. $\dfrac{5}{2}$.
B. 4 .
C. $\dfrac{3}{2}$.
D. 2 .
A. $\dfrac{5}{2}$.
B. 4 .
C. $\dfrac{3}{2}$.
D. 2 .
Đặt $I_1=\int_{\dfrac{\pi}{4}}^{\dfrac{\pi}{2}} \cot x . f\left(\sin ^2 x\right) \mathrm{d} x=1, I_2==\int_1^{16} \dfrac{f(\sqrt{x})}{x} \mathrm{~d} x=1$.
*) Đặt $t=\sin ^2 x \Rightarrow \mathrm{d} t=2 \sin x \cdot \cos x \mathrm{~d} x=2 \sin ^2 x \cdot \cot x \mathrm{~d} x=2 t \cdot \cot x \mathrm{~d} x$.
$
I_1=\int_{\dfrac{\pi}{4}}^{\dfrac{\pi}{2}} \cot x \cdot f\left(\sin ^2 x\right) \mathrm{d} x=\int_{\dfrac{1}{2}}^1 f(t) \cdot \dfrac{1}{2 t} \mathrm{~d} t=\dfrac{1}{2} \int_{\dfrac{1}{2}}^1 \dfrac{f(t)}{t} \mathrm{~d} t=\dfrac{1}{2} \int_{\dfrac{1}{2}}^1 \dfrac{f(x)}{x} \mathrm{~d} x
$
Suy ra $\int_{\dfrac{1}{2}}^1 \dfrac{f(x)}{x} \mathrm{~d} x=2 I_1=2$.
*) Đăt $t=\sqrt{x} \Rightarrow 2 t \mathrm{~d} t=\mathrm{d} x$.
$I_2=\int_1^{16} \dfrac{f(\sqrt{x})}{x} \mathrm{~d} x=\int_1^4 \dfrac{f(t)}{t^2} 2 t \mathrm{~d} t=2 \int_1^4 \dfrac{f(t)}{t} \mathrm{~d} t=2 \int_1^4 \dfrac{f(x)}{x} \mathrm{~d} x$
Suy ra $\int_1^4 \dfrac{f(x)}{x} \mathrm{~d} x=\dfrac{1}{2} I_2=\dfrac{1}{2}$.
*) Đặt $t=4 x \Rightarrow \mathrm{d} t=4 \mathrm{~d} x$.
Ta có : $\int_{\dfrac{1}{8}}^1 \dfrac{f(4 x)}{x} \mathrm{~d} x=\int_{\dfrac{1}{2}}^4 \dfrac{f(t)}{t} \mathrm{dt}=\int_{\dfrac{1}{2}}^1 \dfrac{f(x)}{x} \mathrm{~d} x+\int_1^4 \dfrac{f(x)}{x} \mathrm{~d} x=2+\dfrac{1}{2}=\dfrac{5}{2}$.
*) Đặt $t=\sin ^2 x \Rightarrow \mathrm{d} t=2 \sin x \cdot \cos x \mathrm{~d} x=2 \sin ^2 x \cdot \cot x \mathrm{~d} x=2 t \cdot \cot x \mathrm{~d} x$.
I_1=\int_{\dfrac{\pi}{4}}^{\dfrac{\pi}{2}} \cot x \cdot f\left(\sin ^2 x\right) \mathrm{d} x=\int_{\dfrac{1}{2}}^1 f(t) \cdot \dfrac{1}{2 t} \mathrm{~d} t=\dfrac{1}{2} \int_{\dfrac{1}{2}}^1 \dfrac{f(t)}{t} \mathrm{~d} t=\dfrac{1}{2} \int_{\dfrac{1}{2}}^1 \dfrac{f(x)}{x} \mathrm{~d} x
$
Suy ra $\int_{\dfrac{1}{2}}^1 \dfrac{f(x)}{x} \mathrm{~d} x=2 I_1=2$.
*) Đăt $t=\sqrt{x} \Rightarrow 2 t \mathrm{~d} t=\mathrm{d} x$.
Suy ra $\int_1^4 \dfrac{f(x)}{x} \mathrm{~d} x=\dfrac{1}{2} I_2=\dfrac{1}{2}$.
*) Đặt $t=4 x \Rightarrow \mathrm{d} t=4 \mathrm{~d} x$.
Ta có : $\int_{\dfrac{1}{8}}^1 \dfrac{f(4 x)}{x} \mathrm{~d} x=\int_{\dfrac{1}{2}}^4 \dfrac{f(t)}{t} \mathrm{dt}=\int_{\dfrac{1}{2}}^1 \dfrac{f(x)}{x} \mathrm{~d} x+\int_1^4 \dfrac{f(x)}{x} \mathrm{~d} x=2+\dfrac{1}{2}=\dfrac{5}{2}$.
Đáp án A.