Cho hàm số \(y = f\left( x \right)\) có \(f\left( 0 \right) = 0\) và \(f'\left( x \right) = {\sin ^8}x - {\cos ^8}x - 4{\sin ^6}x,{\rm{ }}\forall x \in \mathbb{R}\). Kết quả của tích phân \(I = \int\limits_0^\pi  {8f\left( x \right){\rm{d}}x} \) được cho dưới dạng \(a{\pi ^2}\). Tìm giá trị của \(a\).

Ta có: \({\sin ^8}x - {\cos ^8}x - 4{\sin ^6}x = \left( {{{\sin }^4}x - {{\cos }^4}x} \right)\left( {{{\sin }^4}x + {{\cos }^4}x} \right) - 4{\sin ^6}x\)

                                                             \( = \left( {{{\sin }^2}x - {{\cos }^2}x} \right)\left( {{{\sin }^4}x + {{\cos }^4}x} \right) - 4{\sin ^6}x\)

                                                             \( = {\cos ^4}x.{\sin ^2}x - {\sin ^4}x.{\cos ^2}x - {\cos ^6}x - 3{\sin ^6}x\)

                                                             \( = {\cos ^4}x.{\sin ^2}x - {\sin ^4}x.{\cos ^2}x - 2{\sin ^6}x - \left( {{{\cos }^6}x + {{\sin }^6}x} \right)\)

                                                             \( = {\sin ^2}x\left( {{{\cos }^4}x - {{\sin }^4}x} \right) - {\sin ^4}x\left( {{{\cos }^2}x + {{\sin }^2}x} \right) - \left( {1 - 3{{\cos }^2}x.{{\sin }^2}x} \right)\)

                                                             \( = 4{\cos ^2}x.{\sin ^2}x - 2{\sin ^4}x - 1\)

                                                             \( =  - \frac{3}{4}\cos 4x + \cos 2x - \frac{5}{4}\).

Do đó \(f\left( x \right) = \int {f'\left( x \right){\rm{d}}x}  = \int {\left( { - \frac{3}{4}\cos 4x + \cos 2x - \frac{5}{4}} \right){\rm{d}}x}  =  - \frac{3}{{16}}\sin 4x + \frac{1}{2}\sin 2x - \frac{5}{4}x + C\)

Vì \(f\left( 0 \right) = 0\) nên \( - \frac{3}{{16}}\sin 4.0 + \frac{1}{2}\sin 2.0 - \frac{5}{4}.0 + C = 0 \Leftrightarrow C = 0\).

Vậy \(f\left( x \right) =  - \frac{3}{{16}}\sin 4x + \frac{1}{2}\sin 2x - \frac{5}{4}x\).

Ta có \(I = \int\limits_0^\pi  {8f\left( x \right){\rm{d}}x} \)

                                    \(\begin{array}{l} = \int\limits_0^\pi  {8\left( { - \frac{3}{{16}}\sin 4x + \frac{1}{2}\sin 2x - \frac{5}{4}x} \right){\rm{d}}x} \\ = \int\limits_0^\pi  {\left( { - \frac{3}{2}\sin 4x + 4\sin 2x - 10x} \right){\rm{d}}x} \\ = \left. {\left( {\frac{3}{8}\cos 4x - 2\cos 2x - 5{x^2}} \right)} \right|_0^\pi  =  - 5{\pi ^2}\end{array}\)