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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $\mathbb{R}$ và thỏa mãn $\sin xf\left( \cos x \right)+\cos xf\left( \sin x \right)=\sin 2x-\dfrac{1}{2}{{\sin }^{3}}2x$ với mọi $x\in \mathbb{R}$. Tính tích phân $I=\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$.
A. $1$.
B. $\dfrac{1}{6}$.
C. $\dfrac{2}{3}$.
D. $\dfrac{1}{3}$.
A. $1$.
B. $\dfrac{1}{6}$.
C. $\dfrac{2}{3}$.
D. $\dfrac{1}{3}$.
Ta có: $\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ \sin xf\left( \cos x \right)+\cos xf\left( \sin x \right) \right]\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \sin 2x-\dfrac{1}{2}{{\sin }^{3}}2x \right)\text{d}x}$
$\Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sin xf\left( \cos x \right)\text{d}x}+\int\limits_{0}^{\dfrac{\pi }{2}}{\cos xf\left( \sin x \right)\text{d}x}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x\left( 1+{{\cos }^{2}}2x \right)\text{d}x}$
* Tính ${{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\sin xf\left( \cos x \right)\text{d}x}$
Đặt $t=\cos x\Rightarrow \text{d}t=-\sin x\text{d}x\Rightarrow -\text{d}t=\sin x\text{d}x$
Đổi cận: $x=0\Rightarrow t=1 ; x=\dfrac{\pi }{2}\Rightarrow t=0$.
Ta có: ${{I}_{1}}=\int\limits_{0}^{1}{f\left( t \right)\text{d}t}=\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$.
* Tương tự , ta tính được: ${{I}_{2}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\cos xf\left( \sin x \right)\text{d}x}=\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$.
* Tính ${{I}_{3}}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x\left( 1+{{\cos }^{2}}2x \right)dx}=-\dfrac{1}{4}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1+{{\cos }^{2}}2x \right)d\left( \cos 2x \right)}$
$=-\dfrac{1}{4}\left. \left( \cos 2x+\dfrac{1}{3}{{\cos }^{3}}2x \right) \right|_{0}^{\dfrac{\pi }{2}}=-\dfrac{1}{4}.\dfrac{-4}{3}+\dfrac{1}{4}.\dfrac{4}{3}=\dfrac{2}{3}$.
Do đó $\int\limits_{0}^{\dfrac{\pi }{2}}{\sin xf\left( \cos x \right)dx}+\int\limits_{0}^{\dfrac{\pi }{2}}{\cos xf\left( \sin x \right)dx}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x\left( 1+{{\cos }^{2}}2x \right)dx}$ trở thành:
$2\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\dfrac{2}{3}\Leftrightarrow \int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\dfrac{1}{3}$.
$\Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sin xf\left( \cos x \right)\text{d}x}+\int\limits_{0}^{\dfrac{\pi }{2}}{\cos xf\left( \sin x \right)\text{d}x}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x\left( 1+{{\cos }^{2}}2x \right)\text{d}x}$
* Tính ${{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\sin xf\left( \cos x \right)\text{d}x}$
Đặt $t=\cos x\Rightarrow \text{d}t=-\sin x\text{d}x\Rightarrow -\text{d}t=\sin x\text{d}x$
Đổi cận: $x=0\Rightarrow t=1 ; x=\dfrac{\pi }{2}\Rightarrow t=0$.
Ta có: ${{I}_{1}}=\int\limits_{0}^{1}{f\left( t \right)\text{d}t}=\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$.
* Tương tự , ta tính được: ${{I}_{2}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\cos xf\left( \sin x \right)\text{d}x}=\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$.
* Tính ${{I}_{3}}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x\left( 1+{{\cos }^{2}}2x \right)dx}=-\dfrac{1}{4}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1+{{\cos }^{2}}2x \right)d\left( \cos 2x \right)}$
$=-\dfrac{1}{4}\left. \left( \cos 2x+\dfrac{1}{3}{{\cos }^{3}}2x \right) \right|_{0}^{\dfrac{\pi }{2}}=-\dfrac{1}{4}.\dfrac{-4}{3}+\dfrac{1}{4}.\dfrac{4}{3}=\dfrac{2}{3}$.
Do đó $\int\limits_{0}^{\dfrac{\pi }{2}}{\sin xf\left( \cos x \right)dx}+\int\limits_{0}^{\dfrac{\pi }{2}}{\cos xf\left( \sin x \right)dx}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x\left( 1+{{\cos }^{2}}2x \right)dx}$ trở thành:
$2\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\dfrac{2}{3}\Leftrightarrow \int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\dfrac{1}{3}$.
Đáp án D.