Tích phân \(I=\int_{0}^{\frac{\pi}{2}}(\cos x-1) \cos ^{2} x d x\) có giá trị là:

* Hướng dẫn giải

Ta có:

\(I=\int_{0}^{\frac{\pi}{2}}(\cos x-1) \cos ^{2} x d x\)

\(=\int_{0}^{\frac{\pi}{2}} \cos x\left(1-\sin ^{2} x\right) d x-\int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x\)

\(=\left.\left(t-\frac{t^{3}}{3}\right)\right|_{0} ^{1}-\left.\frac{1}{2}\left(x+\frac{1}{2} \sin 2 x\right)\right|_{0} ^{\frac{\pi}{2}}\)

\(=\frac{2}{3}-\frac{\pi}{4}\) với \(t=\sin x\)