Đề là cho \(\int\limits^{\dfrac{\pi}{2}}_0sin2x.f\left(cos^2x\right)dx=1\)
Tính \(\int\limits^1_0\left[2f\left(1-x\right)-3x^2+5\right]dx\)
Đúng ko nhỉ?
Xét \(\int\limits^{\dfrac{\pi}{2}}_0sin2x.f\left(cos^2x\right)dx\)
Đặt \(cos^2x=1-u\Rightarrow-2sinx.cosxdx=-du\) \(\Rightarrow sin2xdx=du\)
\(\left\{{}\begin{matrix}x=0\Rightarrow u=0\\x=\dfrac{\pi}{2}\Rightarrow u=1\end{matrix}\right.\) \(\Rightarrow I=\int\limits^1_0f\left(1-u\right)du=\int\limits^1_0f\left(1-x\right)dx\)
\(\Rightarrow\int\limits^1_0f\left(1-x\right)dx=1\)
\(\Rightarrow\int\limits^1_0\left[2f\left(1-x\right)-3x^2+5\right]dx=2\int\limits^1_0f\left(1-x\right)dx-\int\limits^1_0\left(3x^2-5\right)dx\)
\(=2.1-\left(-4\right)=6\)
