Cho hàm số \(f\left( x \right)\) có \(f\left( \frac{\pi }{2} \right)=0\) và \({f}'\left( x \right)=\sin x.{{\sin }^{2}}2x,\forall x\in R\). Khi đó \(\int\limits_{0}^{\frac{\pi }{2}...

* Hướng dẫn giải

Ta có \({f}'\left( x \right)=\sin x.{{\sin }^{2}}2x,\forall x\in R\) nên \(f\left( x \right)\) là một nguyên hàm của \({f}'\left( x \right)\).

Có \(\int{{f}'\left( x \right)\text{d}x}=\int{\sin x.{{\sin }^{2}}2x\text{d}x}=\int{\sin x.\frac{1-\cos 4x}{2}\text{d}x}=\int{\frac{\sin x}{2}\text{d}x-\int{\frac{\sin x.\cos 4x}{2}\text{d}x}}\)

\(=\frac{1}{2}\int{\sin x}\text{d}x-\frac{1}{4}\int{\left( \sin 5x-\sin 3x \right)\text{d}x=-\frac{1}{2}\cos x+\frac{1}{20}\cos 5x-\frac{1}{12}\cos 3x+C}\).

Suy ra \(f\left( x \right)=-\frac{1}{2}\cos x+\frac{1}{20}\cos 5x-\frac{1}{12}\cos 3x+C,\forall x\in \mathbb{R}$. Mà \(f\left( \frac{\pi }{2} \right)=0\Rightarrow C=0\).

Do đó \(f\left( x \right)=-\frac{1}{2}\cos x+\frac{1}{20}\cos 5x-\frac{1}{12}\cos 3x,\forall x\in R\). Khi đó:

\(\int\limits_{0}^{\frac{\pi }{2}}{f\left( x \right)\text{d}x}=\int\limits_{0}^{\frac{\pi }{2}}{\left( -\frac{1}{2}\cos x+\frac{1}{20}\cos 5x-\frac{1}{12}\cos 3x \right)\text{d}x}=\left. \left( -\frac{1}{2}\sin x+\frac{1}{100}\sin 5x-\frac{1}{36}\sin 3x \right) \right|_{0}^{\frac{\pi }{2}}=-\frac{104}{225}\).