Câu hỏi của tyntran1 - Toán lớp 11

\(\int\limits^{\frac{\pi}{2}}_0xcosxdx+\int\limits^{\frac{\pi}{2}}_0sinx.cosxdx=I_1+I_2\)

Tính  \(I_1=\int\limits^{\frac{\pi}{2}}_0x.\left(\sin x\right)'dx=x\sin x-\int\limits^{\frac{\pi}{2}}_0\sin xdx=\frac{\pi}{2}+\cos x\left(0;\frac{\pi}{2}\right)=\frac{\pi}{2}+\cos\frac{\pi}{2}-\cos0=\frac{\pi}{2}-1\)

tính \(I_2=\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x}{2}dx=\left(\frac{-\cos2x}{4}\right)^{\frac{\pi}{2}}_0=-\left(-\cos0\right)=1\)

=> I = \(\frac{\pi}{2}\)