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Câu hỏi: Cho hàm số $f(x)$ có $f(0)=0$ và ${f}'(x)=\sin x{{\sin }^{2}}2x,\forall x\in \mathbb{R}$. Khi đó $\int\limits_{0}^{\pi }{f(x)\text{d}x}$ bằng
A. $\dfrac{7}{30}\pi $.
B. $\dfrac{-7}{30}\pi $.
C. 0.
D. $\dfrac{8}{15}\pi $.
A. $\dfrac{7}{30}\pi $.
B. $\dfrac{-7}{30}\pi $.
C. 0.
D. $\dfrac{8}{15}\pi $.
Ta có: ${f}'(x)=\sin x{{\sin }^{2}}2x=\sin x.{{(2\sin x\cos x)}^{2}}=4\sin x{{\sin }^{2}}x{{\cos }^{2}}x$
$=4\sin x(1-{{\cos }^{2}}x){{\cos }^{2}}x=4\sin x\left( {{\cos }^{2}}x-{{\cos }^{4}}x \right)$, $\forall x\in \mathbb{R}$.
Suy ra: $\int{{f}'(x)\text{d}x}=\int{4\sin x({{\cos }^{2}}x-{{\cos }^{4}}x)\text{d}x}=-4\int{({{\cos }^{2}}x-{{\cos }^{4}}x)\text{d}(\cos x)}$
$=-4\left( \dfrac{{{\cos }^{3}}x}{3}-\dfrac{{{\cos }^{5}}x}{5} \right)+C$ $=\dfrac{4}{5}{{\cos }^{5}}x-\dfrac{4}{3}{{\cos }^{3}}x+C$.
Do đó: $f(x)=\dfrac{4}{5}{{\cos }^{5}}x-\dfrac{4}{3}{{\cos }^{3}}x+C$, $\forall x\in \mathbb{R}$.
Vì $f(0)=0$ nên $\dfrac{4}{5}{{\cos }^{5}}0-\dfrac{4}{3}{{\cos }^{3}}0+C=0$, hay $C=\dfrac{8}{15}$.
Vậy $f(x)=\dfrac{4}{5}{{\cos }^{5}}x-\dfrac{4}{3}{{\cos }^{3}}x+\dfrac{8}{15},\forall x\in \mathbb{R}$.
Ta có: $\int\limits_{0}^{\pi }{f(x)\text{d}x}=\int\limits_{0}^{\pi }{\left( \dfrac{4}{5}{{\cos }^{5}}x-\dfrac{4}{3}{{\cos }^{3}}x+\dfrac{8}{15} \right)\text{d}x}=4\int\limits_{0}^{\pi }{\cos x\left( \dfrac{{{\cos }^{4}}x}{5}-\dfrac{{{\cos }^{2}}x}{3} \right)\text{d}x+\int\limits_{0}^{\pi }{\dfrac{8}{15}\text{d}x}}$
$=4\int\limits_{0}^{\pi }{\cos x\left( \dfrac{{{(1-{{\sin }^{2}}x)}^{2}}}{5}-\dfrac{1-{{\sin }^{2}}x}{3} \right)\text{d}x+\left( \dfrac{8}{15}x \right)\left| _{0}^{\pi } \right.}=4\int\limits_{0}^{\pi }{\left( \dfrac{{{\sin }^{4}}x}{5}-\dfrac{{{\sin }^{2}}x}{15}-\dfrac{2}{15} \right)}\cos x\text{d}x+\dfrac{8\pi }{15}$
$=4\int\limits_{0}^{\pi }{\left( \dfrac{{{\sin }^{4}}x}{5}-\dfrac{{{\sin }^{2}}x}{15}-\dfrac{2}{15} \right)\text{d}\left( \sin x \right)}+\dfrac{8\pi }{15}=4\left( \dfrac{{{\sin }^{5}}x}{25}-\dfrac{{{\sin }^{3}}x}{45}-\dfrac{2\sin x}{15} \right)\left| _{0}^{\pi } \right.+\dfrac{8\pi }{15}$ $=\dfrac{8\pi }{15}$.
$=4\sin x(1-{{\cos }^{2}}x){{\cos }^{2}}x=4\sin x\left( {{\cos }^{2}}x-{{\cos }^{4}}x \right)$, $\forall x\in \mathbb{R}$.
Suy ra: $\int{{f}'(x)\text{d}x}=\int{4\sin x({{\cos }^{2}}x-{{\cos }^{4}}x)\text{d}x}=-4\int{({{\cos }^{2}}x-{{\cos }^{4}}x)\text{d}(\cos x)}$
$=-4\left( \dfrac{{{\cos }^{3}}x}{3}-\dfrac{{{\cos }^{5}}x}{5} \right)+C$ $=\dfrac{4}{5}{{\cos }^{5}}x-\dfrac{4}{3}{{\cos }^{3}}x+C$.
Do đó: $f(x)=\dfrac{4}{5}{{\cos }^{5}}x-\dfrac{4}{3}{{\cos }^{3}}x+C$, $\forall x\in \mathbb{R}$.
Vì $f(0)=0$ nên $\dfrac{4}{5}{{\cos }^{5}}0-\dfrac{4}{3}{{\cos }^{3}}0+C=0$, hay $C=\dfrac{8}{15}$.
Vậy $f(x)=\dfrac{4}{5}{{\cos }^{5}}x-\dfrac{4}{3}{{\cos }^{3}}x+\dfrac{8}{15},\forall x\in \mathbb{R}$.
Ta có: $\int\limits_{0}^{\pi }{f(x)\text{d}x}=\int\limits_{0}^{\pi }{\left( \dfrac{4}{5}{{\cos }^{5}}x-\dfrac{4}{3}{{\cos }^{3}}x+\dfrac{8}{15} \right)\text{d}x}=4\int\limits_{0}^{\pi }{\cos x\left( \dfrac{{{\cos }^{4}}x}{5}-\dfrac{{{\cos }^{2}}x}{3} \right)\text{d}x+\int\limits_{0}^{\pi }{\dfrac{8}{15}\text{d}x}}$
$=4\int\limits_{0}^{\pi }{\cos x\left( \dfrac{{{(1-{{\sin }^{2}}x)}^{2}}}{5}-\dfrac{1-{{\sin }^{2}}x}{3} \right)\text{d}x+\left( \dfrac{8}{15}x \right)\left| _{0}^{\pi } \right.}=4\int\limits_{0}^{\pi }{\left( \dfrac{{{\sin }^{4}}x}{5}-\dfrac{{{\sin }^{2}}x}{15}-\dfrac{2}{15} \right)}\cos x\text{d}x+\dfrac{8\pi }{15}$
$=4\int\limits_{0}^{\pi }{\left( \dfrac{{{\sin }^{4}}x}{5}-\dfrac{{{\sin }^{2}}x}{15}-\dfrac{2}{15} \right)\text{d}\left( \sin x \right)}+\dfrac{8\pi }{15}=4\left( \dfrac{{{\sin }^{5}}x}{25}-\dfrac{{{\sin }^{3}}x}{45}-\dfrac{2\sin x}{15} \right)\left| _{0}^{\pi } \right.+\dfrac{8\pi }{15}$ $=\dfrac{8\pi }{15}$.
Đáp án D.