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Câu hỏi: Cho hàm số $f\left(x \right)$ biết $f\left(\pi \right)=0$ và $f'\left(x \right)=2\sin x-3{{\sin }^{3}}x,\forall x\in \mathbb{R}$, biết $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f\left(x \right)}{{{\sin }^{2}}x+1}dx}=a-\dfrac{b\pi }{c}; a, b, c$ là các số nguyên dương và $\dfrac{b}{c}$ tối giản. Tổng $S=a+b+c$ bằng:
A. $6$.
B. $5$.
C. $8$.
D. $7$.
A. $6$.
B. $5$.
C. $8$.
D. $7$.
Ta có :
$\begin{aligned}
& f\left(x \right)=\int{\left(2\sin x-3{{\sin }^{3}}x \right)dx}=\int{\sin x\left(2-3{{\sin }^{2}}x \right)dx}=\int{\sin x\left(3{{\cos }^{2}}x-1 \right)dx} \\
& =-\int{\left(3{{\cos }^{2}}x-1 \right)d\left(\cos x \right)}=-{{\cos }^{3}}x+\cos x+C \\
\end{aligned}$
Vì $f\left(\pi \right)=0$ nên $-co{{s}^{3}}\pi +\cos \pi +C=0\Leftrightarrow C=0$. Vậy $f\left(x \right)=-co{{s}^{3}}x+\cos x$ Xét $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f\left(x \right)}{{{\sin }^{2}}x+1}dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x-co{{s}^{3}}x}{{{\sin }^{2}}x+1}dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x\left(1-co{{s}^{2}}x \right)}{{{\sin }^{2}}x+1}dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x.{{\sin }^{2}}x}{{{\sin }^{2}}x+1}dx}$
Cách 1: Đặt $\sin x=u; du=\cos xdx;$
Đổi cận: $x=0\Rightarrow u=0; x=\dfrac{\pi }{2}\Rightarrow u=1.$
$I=\int\limits_{0}^{1}{\dfrac{{{u}^{2}}}{{{u}^{2}}+1}du}=\int\limits_{0}^{1}{\left(1-\dfrac{1}{{{u}^{2}}+1} \right)du=\left. U \right|_{0}^{1}}-\int\limits_{0}^{1}{\dfrac{1}{{{u}^{2}}+1}du}.$
Xét $J=\int\limits_{0}^{1}{\dfrac{1}{{{u}^{2}}+1}}du,$ đặt $u=\tan t, t\in \left(0;\dfrac{\pi }{2} \right); du=\dfrac{1}{co{{s}^{2}}t}dt=\left({{\tan }^{2}}t+1 \right)dt.$
Đổi cận $u=0\Rightarrow t=0; u=1\Rightarrow t=\dfrac{\pi }{4}.$
$J=\int\limits_{0}^{1}{\dfrac{1}{{{u}^{2}}+1}}du=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{{{\tan }^{2}}t+1}{{{\tan }^{2}}t+1}}dt=\left. T \right|_{0}^{\dfrac{\pi }{4}}=\dfrac{\pi }{4}.$
Vậy $I=1-J=1-\dfrac{\pi }{4}.$
Cách 2: Đặt $\sin x=\tan t, t\in \left(0;\dfrac{\pi }{2} \right).$ Lấy vi phân 2 vế, ta có $\cos xdx=\left({{\tan }^{2}}t+1 \right)dt;$
Đổi cận $x=0\Rightarrow t=0; x=\dfrac{\pi }{2}\Rightarrow t=\dfrac{\pi }{4}.$
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x{{\sin }^{2}}x}{{{\sin }^{2}}x+1}dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{{{\tan }^{2}}t}{{{\tan }^{2}}t+1}\left({{\tan }^{2}}t+1 \right)dt}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left(\dfrac{1}{co{{s}^{2}}t}-1 \right)dt}=\left. \left(\tan t-t \right) \right|_{0}^{\dfrac{\pi }{4}}=1-\dfrac{\pi }{4}.$
Vậy $S=a+b+c=6.$
$\begin{aligned}
& f\left(x \right)=\int{\left(2\sin x-3{{\sin }^{3}}x \right)dx}=\int{\sin x\left(2-3{{\sin }^{2}}x \right)dx}=\int{\sin x\left(3{{\cos }^{2}}x-1 \right)dx} \\
& =-\int{\left(3{{\cos }^{2}}x-1 \right)d\left(\cos x \right)}=-{{\cos }^{3}}x+\cos x+C \\
\end{aligned}$
Vì $f\left(\pi \right)=0$ nên $-co{{s}^{3}}\pi +\cos \pi +C=0\Leftrightarrow C=0$. Vậy $f\left(x \right)=-co{{s}^{3}}x+\cos x$ Xét $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f\left(x \right)}{{{\sin }^{2}}x+1}dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x-co{{s}^{3}}x}{{{\sin }^{2}}x+1}dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x\left(1-co{{s}^{2}}x \right)}{{{\sin }^{2}}x+1}dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x.{{\sin }^{2}}x}{{{\sin }^{2}}x+1}dx}$
Cách 1: Đặt $\sin x=u; du=\cos xdx;$
Đổi cận: $x=0\Rightarrow u=0; x=\dfrac{\pi }{2}\Rightarrow u=1.$
$I=\int\limits_{0}^{1}{\dfrac{{{u}^{2}}}{{{u}^{2}}+1}du}=\int\limits_{0}^{1}{\left(1-\dfrac{1}{{{u}^{2}}+1} \right)du=\left. U \right|_{0}^{1}}-\int\limits_{0}^{1}{\dfrac{1}{{{u}^{2}}+1}du}.$
Xét $J=\int\limits_{0}^{1}{\dfrac{1}{{{u}^{2}}+1}}du,$ đặt $u=\tan t, t\in \left(0;\dfrac{\pi }{2} \right); du=\dfrac{1}{co{{s}^{2}}t}dt=\left({{\tan }^{2}}t+1 \right)dt.$
Đổi cận $u=0\Rightarrow t=0; u=1\Rightarrow t=\dfrac{\pi }{4}.$
$J=\int\limits_{0}^{1}{\dfrac{1}{{{u}^{2}}+1}}du=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{{{\tan }^{2}}t+1}{{{\tan }^{2}}t+1}}dt=\left. T \right|_{0}^{\dfrac{\pi }{4}}=\dfrac{\pi }{4}.$
Vậy $I=1-J=1-\dfrac{\pi }{4}.$
Cách 2: Đặt $\sin x=\tan t, t\in \left(0;\dfrac{\pi }{2} \right).$ Lấy vi phân 2 vế, ta có $\cos xdx=\left({{\tan }^{2}}t+1 \right)dt;$
Đổi cận $x=0\Rightarrow t=0; x=\dfrac{\pi }{2}\Rightarrow t=\dfrac{\pi }{4}.$
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x{{\sin }^{2}}x}{{{\sin }^{2}}x+1}dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{{{\tan }^{2}}t}{{{\tan }^{2}}t+1}\left({{\tan }^{2}}t+1 \right)dt}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left(\dfrac{1}{co{{s}^{2}}t}-1 \right)dt}=\left. \left(\tan t-t \right) \right|_{0}^{\dfrac{\pi }{4}}=1-\dfrac{\pi }{4}.$
Vậy $S=a+b+c=6.$
Đáp án A.