Cho hàm số \(f\left( x \right)\) có \(f\left( 0 \right)=\frac{1}{8}\) và \({f}'\left( x \right)=x{{\cos }^{2}}x, \forall x\in \mathbb{R}\). Tích phân \(\int\limits_{\frac{\pi }{2}}...

* Hướng dẫn giải

Ta có \(\int{x{{\cos }^{2}}x\text{d}x}=\int{x.\frac{1+\cos 2x}{2}}\text{d}x=\int{\frac{x}{2}\text{d}x+\frac{1}{2}\int{x\cos 2x\text{d}x}}=\frac{{{x}^{2}}}{4}+\frac{1}{4}\int{x\text{d}\left( \sin 2x \right)}\)

\(=\frac{{{x}^{2}}}{4}+\frac{1}{4}x\sin 2x-\frac{1}{4}\int{\sin 2x\text{d}x}=\frac{{{x}^{2}}}{4}+\frac{1}{4}x\sin 2x+\frac{1}{8}\cos 2x+C\)

Suy ra \(f\left( x \right)=\frac{{{x}^{2}}}{4}+\frac{1}{4}x\sin 2x+\frac{1}{8}\cos 2x+C\)

Mà \(f\left( 0 \right)=\frac{1}{8}\Rightarrow \frac{1}{8}+C=\frac{1}{8}\Leftrightarrow C=0\)

Do đó \(f\left( x \right)=\frac{{{x}^{2}}}{4}+\frac{1}{4}x\sin 2x+\frac{1}{8}\cos 2x\).

Ta có \(\int\limits_{\frac{\pi }{2}}^{\pi }{\frac{8f\left( x \right)-\cos 2x}{x}\text{d}x}=\int\limits_{\frac{\pi }{2}}^{\pi }{\left( 2x+2\sin 2x \right)\text{d}x}=\left. \left( {{x}^{2}}-\cos 2x \right) \right|_{\frac{\pi }{2}}^{\pi }={{\pi }^{2}}-1-\frac{{{\pi }^{2}}}{4}-1=\frac{3{{\pi }^{2}}-8}{4}\).