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Câu hỏi: Họ nguyên hàm của hàm số $f\left( x \right)=\cos x-{{\cos }^{3}}x$ là
A. $\mathop{\int }^{}f\left( x \right)dx=-\dfrac{1}{3}{{\sin }^{3}}x+C.$
B. $\mathop{\int }^{}f\left( x \right)dx=\sin x-\dfrac{1}{4}{{\sin }^{4}}x+C.$
C. $\mathop{\int }^{}f\left( x \right)dx=\dfrac{1}{3}{{\sin }^{3}}x+C.$
D. $\mathop{\int }^{}f\left( x \right)dx=-\sin x+\dfrac{1}{4}{{\sin }^{4}}x+C.$
A. $\mathop{\int }^{}f\left( x \right)dx=-\dfrac{1}{3}{{\sin }^{3}}x+C.$
B. $\mathop{\int }^{}f\left( x \right)dx=\sin x-\dfrac{1}{4}{{\sin }^{4}}x+C.$
C. $\mathop{\int }^{}f\left( x \right)dx=\dfrac{1}{3}{{\sin }^{3}}x+C.$
D. $\mathop{\int }^{}f\left( x \right)dx=-\sin x+\dfrac{1}{4}{{\sin }^{4}}x+C.$
Ta có $f\left( x \right)=\left( 1-{{\cos }^{2}}x \right)\cos x={{\sin }^{2}}x.\cos x$
Suy ra $\mathop{\int }^{}f\left( x \right)dx=\mathop{\int }^{}{{\sin }^{2}}x.\cos xdx=\mathop{\int }^{}{{\sin }^{2}}xd\left( \sin x \right)=\dfrac{{{\sin }^{3}}x}{3}+C.$ Chọn C
Suy ra $\mathop{\int }^{}f\left( x \right)dx=\mathop{\int }^{}{{\sin }^{2}}x.\cos xdx=\mathop{\int }^{}{{\sin }^{2}}xd\left( \sin x \right)=\dfrac{{{\sin }^{3}}x}{3}+C.$ Chọn C
Đáp án C.