Câu hỏi của oOoLEOoOO

\(I=\int_0^{\dfrac{\pi}{2}}\left(cos2x+sin^2x\right)cosx.dx\)\(=\int_0^{\dfrac{\pi}{2}}\left(cos^2x-sin^2x+sin^2x\right)cosx.dx\)\(=\int_0^{\dfrac{\pi}{2}}cos^2x.cosx.dx\)Đặt t=sinx, ta có: \(dt=cosx.dx\)\(\Rightarrow I=\int_0^1\left(1-t^2\right)dt\)\(=t-\dfrac{t^3}{3}|_0^1\)\(=1-\dfrac{1}{3}=\dfrac{2}{3}\)Câu trả lời: \(I=\int_0^{\dfrac{\pi}{2}}\left(cos2x+sin^2x\right)cosx.dx\)\(=\int_0^{\dfrac{\pi}{2}}\left(cos^2x-sin^2x+sin^2x\right)cosx.dx\)\(=\int_0^{\dfrac{\pi}{2}}cos^2x.cosx.dx\)Đặt t=sinx, ta có: \(dt=cosx.dx\)\(\Rightarrow I=\int_0^1\left(1-t^2\right)dt\)\(=t-\dfrac{t^3}{3}|_0^1\)\(=1-\dfrac{1}{3}=\dfrac{2}{3}\)