Tích phân \(\int\limits^1_0\sqrt{1-x^2}\text{d}x\) bằng
\(\dfrac{\pi}{3}\).\(\dfrac{\pi}{4}\).\(\dfrac{\pi}{6}\).\(1\).Hướng dẫn giải:Vì \(0\le x\le1\) nên ta có thể đặt \(x=\sin t\Rightarrow\) \(\text{d}x=\cos t\text{d}t\)
Đổi cận: \(x|^1_0\Rightarrow t|^{\frac{\pi}{2}}_0\)
\(\int\limits^{\frac{\pi}{2}}_0\sqrt{1-\sin^2t}\cos t.\text{d}t\)
\(=\int\limits^{\frac{\pi}{2}}_0\cos t.\cos t\text{d}t\)
\(=\int\limits^{\frac{\pi}{2}}_0\cos^2t\text{d}t\)
\(=\int\limits^{\frac{\pi}{2}}_0\dfrac{1+\cos2t}{2}\text{d}t\)
\(=\dfrac{1}{2}\int\limits^{\frac{\pi}{2}}_0\text{d}t+\dfrac{1}{2}.\dfrac{1}{2}\int\limits^{\frac{\pi}{2}}_0\cos\left(2t\right)\text{d}\left(2t\right)\)
\(=\left[\dfrac{t}{2}+\dfrac{1}{4}\sin\left(2t\right)\right]|^{\frac{\pi}{2}}_0\)
\(=\dfrac{\pi}{4}\)