Tính \(\displaystyle \int\limits_0^{\frac{\pi }{4}} {\cos 2x} .{\cos ^2}xdx\)
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Lời giải:
Báo saiTa có: \(\displaystyle {\cos ^2}x = \frac{{1 + \cos 2x}}{2}\) \(\displaystyle \Rightarrow \cos 2x.{\cos ^2}x = \frac{1}{2}\cos 2x\left( {1 + \cos 2x} \right)\)
\(\displaystyle = \frac{1}{2}\cos 2x + \frac{1}{2}{\cos ^2}2x\) \(\displaystyle = \frac{1}{2}\cos 2x + \frac{1}{4}\left( {1 + \cos 4x} \right)\) \(\displaystyle = \frac{1}{2}\cos 2x + \frac{1}{4}\cos 4x + \frac{1}{4}\)
Suy ra \(\displaystyle \int\limits_0^{\frac{\pi }{4}} {\cos 2x} .{\cos ^2}xdx\)\(\displaystyle = \int\limits_0^{\frac{\pi }{4}} {\left( {\frac{1}{2}\cos 2x + \frac{1}{4}\cos 4x + \frac{1}{4}} \right)dx} \) \(\displaystyle = \left. {\left( {\frac{1}{4}\sin 2x + \frac{1}{{16}}\sin 4x + \frac{1}{4}x} \right)} \right|_0^{\frac{\pi }{4}}\) \(\displaystyle = \frac{1}{4} + \frac{\pi }{{16}}\)