Cho hàm số f(x), có \(f\lef...

A

- Hướng dẫn giải

Ta có \(I = \int {f'\left( x \right)} dx = \int {\sin x.{{\cos }^2}2xdx = \int {\sin x{{\left( {2{{\cos }^2}x - 1} \right)}^2}dx} } \)

Đặt \(t = \cos x \Rightarrow dt = - \sin xdx\)

Suy ra \(I = - \int {{{\left( {2{t^2} - 1} \right)}^2}dt = - \int {\left( {4{t^4} - 4{t^2} + 1} \right)dt = - \frac{4}{5}{t^5} + \frac{4}{3}{t^3} - t + c} } \)

Hay \(I = - \frac{4}{5}{\cos ^5}x + \frac{4}{3}{\cos ^3}x - \cos x + C \Rightarrow f\left( x \right) = - \frac{4}{5}{\cos ^5}x + \frac{4}{3}{\cos ^3}x - \cos x + C\)

Mà \(f\left( {\frac{\pi }{2}} \right) = 0 \Rightarrow C = 0\). Vậy \(f\left( x \right) = - \frac{4}{5}{\cos ^5}x + \frac{4}{3}{\cos ^3}x - \cos x\)

Tích phân \(J = \int\limits_0^{\frac{\pi }{2}} {f\left( x \right)} dx = \int\limits_0^{\frac{\pi }{2}} {\left( { - \frac{4}{5}{{\cos }^5}x + \frac{4}{3}{{\cos }^3}x - \cos x} \right)} dx\)

\(= \int\limits_0^{\frac{\pi }{2}} {\cos x\left( { - \frac{4}{5}{{\cos }^4}x + \frac{4}{3}{{\cos }^2}x - 1} \right)} dx\)

\( = \int\limits_0^{\frac{\pi }{2}} {\cos x\left( { - \frac{4}{5}{{\left( {1 - {{\sin }^2}x} \right)}^2} + \frac{4}{3}\left( {1 - {{\sin }^2}x} \right) - 1} \right)} dx\)

Đặt \(t = \sin x \Rightarrow dt = \cos xdx\)

Đổi cận \(x = 0 \Rightarrow t = 0;\,\,x = \frac{\pi }{2} \Rightarrow t = 1\)

Khi đó \(J = \int\limits_0^1 {\left[ { - \frac{4}{5}{{\left( {1 - {t^2}} \right)}^2} + \frac{4}{3}\left( {1 - {t^2}} \right) - 1} \right]dt} = - \frac{{121}}{{225}}\)