\(\Rightarrow I=\int_{-1}^{1}(t+1)\sqrt{1-t^2}dt=\int_{-1}^{1}t\sqrt{1-t^2}dt+\int_{-1}^{1}\sqrt{1-t^2}dt\) Tính \(A=\int_{-1}^{1}t\sqrt{1-t^2}dt=-\frac{1}{2}\int_{-1}^{1}(1-t^2)^{\frac{1}{2}}d(1-t^2)=0\) \(B=\int_{-1}^{1}\sqrt{1-t^2}dt\) Đặt \(t =sin u \Rightarrow dt =cosu \ du\)
