有关三角函数的定积分
设\(f(x)\)在\([0,1]\)上连续,则\(\displaystyle\int_{0}^{\pi/2} f(\sin x)=\int_{0}^{\pi/2} f(\cos x)\)
\(I_n=\displaystyle\int_{0}^{\pi/2}\sin^n x=\int_{0}^{\pi/2}\cos^n x,I_n=\frac{n-1}{n}I_{n-2},I_1=1,I_0=\frac{\pi}{2}\)
设\(f(x)\)在\([0,1]\)上连续,则\(\displaystyle\int_{0}^{\pi} f(\sin x)=2\int_{0}^{\pi/2} f(\sin x)\)
设\(f(x)\)在\([0,1]\)上连续,则\(\displaystyle\int_{0}^{\pi} xf(\sin x)=\frac{\pi}{2}\int_{0}^{\pi/2} f(\sin x)\)