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Câu hỏi: Cho hàm số $f\left( x \right)$ có $f\left( \dfrac{\pi }{2} \right)=\dfrac{8}{15}$ và ${f}'\left( x \right)=\cos x.{{\sin }^{2}}2x,\forall \in R$. Khi đó $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}$ bằng:
A. $\dfrac{102}{225}$.
B. $\dfrac{121}{225}$.
C. $\dfrac{104}{225}$.
D. $\dfrac{109}{225}$.
A. $\dfrac{102}{225}$.
B. $\dfrac{121}{225}$.
C. $\dfrac{104}{225}$.
D. $\dfrac{109}{225}$.
Ta có: ${f}'\left( x \right)=\cos x.{{\sin }^{2}}2x=\cos x.{{\left( 2\sin x.\cos x \right)}^{2}}=4\cos x.{{\sin }^{2}}x.{{\cos }^{2}}x=4\cos x.{{\sin }^{2}}x.\left( 1-{{\sin }^{2}}x \right)$
$\Rightarrow f\left( x \right)=\int{f'\left( x \right)}dx=\int{4\cos x.{{\sin }^{2}}x.\left( 1-{{\sin }^{2}}x \right)dx}$. Đặt $t=\sin x\Rightarrow dt=\cos xdx$
Ta có: $I=\int{4{{t}^{2}}\left( 1-{{t}^{2}} \right)}dt=\int{\left( 4{{t}^{2}}-4{{t}^{4}} \right)}dt=\dfrac{4}{3}{{t}^{3}}-\dfrac{4}{5}{{t}^{5}}+c\Rightarrow f\left( x \right)=\dfrac{4}{3}{{\sin }^{3}}x-\dfrac{4}{5}{{\sin }^{5}}x+c$
Vì $f\left( \dfrac{\pi }{2} \right)=\dfrac{8}{15}\Rightarrow C=0\Rightarrow f\left( x \right)=\dfrac{4}{3}{{\sin }^{3}}x-\dfrac{4}{5}{{\sin }^{5}}x$
Vậy $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{4}{3}{{\sin }^{3}}x-\dfrac{4}{5}{{\sin }^{5}}x \right)\text{d}x}=\dfrac{104}{225}$
$\Rightarrow f\left( x \right)=\int{f'\left( x \right)}dx=\int{4\cos x.{{\sin }^{2}}x.\left( 1-{{\sin }^{2}}x \right)dx}$. Đặt $t=\sin x\Rightarrow dt=\cos xdx$
Ta có: $I=\int{4{{t}^{2}}\left( 1-{{t}^{2}} \right)}dt=\int{\left( 4{{t}^{2}}-4{{t}^{4}} \right)}dt=\dfrac{4}{3}{{t}^{3}}-\dfrac{4}{5}{{t}^{5}}+c\Rightarrow f\left( x \right)=\dfrac{4}{3}{{\sin }^{3}}x-\dfrac{4}{5}{{\sin }^{5}}x+c$
Vì $f\left( \dfrac{\pi }{2} \right)=\dfrac{8}{15}\Rightarrow C=0\Rightarrow f\left( x \right)=\dfrac{4}{3}{{\sin }^{3}}x-\dfrac{4}{5}{{\sin }^{5}}x$
Vậy $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{4}{3}{{\sin }^{3}}x-\dfrac{4}{5}{{\sin }^{5}}x \right)\text{d}x}=\dfrac{104}{225}$
Đáp án C.