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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm là $f'\left( x \right)=\left( 1+3{{\cos }^{2}}x \right)\sin x,\forall x\in \mathbb{R}$ và $f\left( 0 \right)=- 4$. Tính $\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ f\left( x \right)+2 \right]} dx$.
A. $\dfrac{5}{3}$.
B. $\dfrac{2}{3}$.
C. $-\dfrac{5}{3}$.
D. $-\dfrac{2}{3}$.
A. $\dfrac{5}{3}$.
B. $\dfrac{2}{3}$.
C. $-\dfrac{5}{3}$.
D. $-\dfrac{2}{3}$.
Ta có: $f\left( x \right)=\int{f'\left( x \right)dx=\int{\left( 1+3{{\cos }^{2}}x \right)\sin xdx}}$
$=-\int{\left( 1+3{{\cos }^{2}}x \right)d\left( \cos x \right)}=-\cos x-{{\cos }^{3}}x+C$
Với $f\left( 0 \right)=- 4\Rightarrow -\cos 0-{{\cos }^{3}}0+C=- 4\Rightarrow C=-2$
Vậy $f\left( x \right)=-\cos x-{{\cos }^{3}}x-2$
Ta có : $\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ f\left( x \right)+2 \right]} dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ -\cos x-{{\cos }^{3}}x \right]} dx=-\dfrac{5}{3}$.
$=-\int{\left( 1+3{{\cos }^{2}}x \right)d\left( \cos x \right)}=-\cos x-{{\cos }^{3}}x+C$
Với $f\left( 0 \right)=- 4\Rightarrow -\cos 0-{{\cos }^{3}}0+C=- 4\Rightarrow C=-2$
Vậy $f\left( x \right)=-\cos x-{{\cos }^{3}}x-2$
Ta có : $\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ f\left( x \right)+2 \right]} dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ -\cos x-{{\cos }^{3}}x \right]} dx=-\dfrac{5}{3}$.
Đáp án C.