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Câu hỏi: Giá trị $\int\limits_{0}^{\dfrac{\pi }{2}}{{{\left( 1-\cos x \right)}^{n}} \sin x \text{d}x}$ bằng
A. $\dfrac{1}{n-1}$.
B. $\dfrac{1}{2n}$.
C. $-\dfrac{1}{n+1}$.
D. $\dfrac{1}{n+1}$.
A. $\dfrac{1}{n-1}$.
B. $\dfrac{1}{2n}$.
C. $-\dfrac{1}{n+1}$.
D. $\dfrac{1}{n+1}$.
Ta có $\int\limits_{0}^{\dfrac{\pi }{2}}{{{\left( 1-\cos x \right)}^{n}} \sin x \text{d}x}=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\left( 1-\cos x \right)}^{n}} d\left( 1-\cos x \right) }=\dfrac{1}{n+1}{{\left( 1-\cos x \right)}^{n+1}}|_{0}^{\dfrac{\pi }{2}}=$ $\dfrac{1}{n+1}$.
Đáp án D.