\(\int\limits^{\frac{\pi}{3}}_0\cos^nx\sin x\text{d}x=\frac{15}{64}\) khi
\(n=1\). \(n=2\). \(n=3\). \(n=4\). Hướng dẫn giải:\(\int\limits^{\frac{\pi}{3}}_0\cos^nx\sin x\text{d}x=-\int\limits^{\frac{\pi}{3}}_0\cos^nx\text{d}\left(\cos x\right)\)
\(=-\dfrac{\cos^{n+1}x}{n+1}|^{\frac{\pi}{3}}_0=-\left(\dfrac{\left(\frac{1}{2}\right)^{n+1}}{n+1}-\dfrac{1}{n+1}\right)\)
\(=\dfrac{1}{n+1}-\dfrac{1}{2^{n+1}\left(n+1\right)}=\dfrac{2^{n+1}-1}{2^{n+1}\left(n+1\right)}\)
Theo yêu cầu ta có:
\(\dfrac{2^{n+1}-1}{2^{n+1}\left(n+1\right)}=\dfrac{15}{64}\)
\(\Rightarrow n=3.\)