Cho hàm số f(x) có f(0) = 0 và

C

- Hướng dẫn giải

Ta có \(f'\left( x \right) = \cos \left( {x + \frac{\pi }{4}} \right){\cos ^2}\left( {2x + \frac{\pi }{2}} \right),\forall x \in R\) nên f(x) là một nguyên hàm của f'(x).

\(\int {f'\left( x \right){\rm{d}}x} = \int {\cos \left( {x + \frac{\pi }{4}} \right){{\cos }^2}\left( {2x + \frac{\pi }{2}} \right){\rm{d}}x = } \int {\cos \left( {x + \frac{\pi }{4}} \right){{\cos }^2}\left( {2\left( {x + \frac{\pi }{4}} \right)} \right){\rm{d}}x} \)

\(= \int {\cos \left( {x + \frac{\pi }{4}} \right)\left( {1 - 2{{\sin }^2}\left( {x + \frac{\pi }{4}} \right)} \right){\rm{d}}x} = I\)

Đặt \(t = \sin \left( {x + \frac{\pi }{4}} \right) \Rightarrow {\rm{d}}t = \cos \left( {x + \frac{\pi }{4}} \right){\rm{d}}x\)

Ta có \(I = \int {\left( {1 - 2{t^2}} \right){\rm{d}}t = t - \frac{2}{3}{t^3} + c = \sin \left( {x + \frac{\pi }{4}} \right) - \frac{2}{3}{{\sin }^3}\left( {x + \frac{\pi }{4}} \right) + C} \)

\(\begin{array}{l}f\left( { - \frac{\pi }{4}} \right) = 0 \Rightarrow C = 0 \Rightarrow f\left( x \right) = \sin \left( {x + \frac{\pi }{4}} \right) - \frac{2}{3}{\sin ^3}\left( {x + \frac{\pi }{4}} \right)\\\int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {f\left( x \right){\rm{d}}x} = \int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\left( {\sin \left( {x + \frac{\pi }{4}} \right) - \frac{2}{3}{{\sin }^3}\left( {x + \frac{\pi }{4}} \right)} \right){\rm{d}}x} \\ = \int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\sin \left( {x + \frac{\pi }{4}} \right){\rm{d}}x} - \frac{2}{3}\int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\sin \left( {x + \frac{\pi }{4}} \right)\left( {1 - {{\cos }^2}\left( {x + \frac{\pi }{4}} \right)} \right)} {\rm{d}}x\\ = \left. { - \cos \left( {x + \frac{\pi }{4}} \right)} \right|_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} + \frac{2}{3}\int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\left( {1 - {{\cos }^2}\left( {x + \frac{\pi }{4}} \right)} \right){\rm{d}}\left( {\cos \left( {x + \frac{\pi }{4}} \right)} \right)} \\ = 1 + \frac{2}{3}\left. {\left( {\cos \left( {x + \frac{\pi }{4}} \right) - \frac{1}{3}{{\cos }^3}\left( {x + \frac{\pi }{4}} \right)} \right)} \right|_{\frac{{ - \pi }}{4}}^{\frac{\pi }{4}} = 1 + \frac{2}{3}\left( { - 1 + \frac{1}{3}} \right) = \frac{5}{9}\end{array}\)