Cho hàm số . Tích phân \(I = \int\limits_0^\pi {\sin 2x.f\left( {{\rm{cos}}x} \right){\rm{d}}x} \) bằng

* Hướng dẫn giải

Do \(\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( 0 \right)=-2\) nên hàm số \(f\left( x \right)\)

Đặt \(t=\cos x\Rightarrow \text{d}t=-\sin x\text{d}x\)

Đổi cận: \(x=0\Rightarrow t=1; x=\pi \Rightarrow t=-1\).

Ta có:

\(\int\limits_{0}^{\pi }{\sin 2x.f\left( \text{cos}x \right)\text{d}x}=\int\limits_{0}^{\pi }{2\sin x.\text{cos}x.f\left( \text{cos}x \right)\text{d}x=-\int\limits_{1}^{-1}{2t.f\left( t \right)\text{d}t}}=2\int\limits_{-1}^{1}{t.f\left( t \right)\text{d}t}\)

\(=2\int\limits_{-1}^{0}{x.f\left( x \right)\text{d}x}+2\int\limits_{0}^{1}{x.f\left( x \right)\text{d}x=2\int\limits_{0}^{1}{x\left( {{x}^{2}}+4x-2 \right)\text{d}x}+2\int\limits_{-1}^{0}{x.\left( 2x-2 \right)\text{d}x}}\)

\( = 2\left( {\frac{{{x^4}}}{4} + \frac{{4{x^3}}}{3} - {x^2}} \right)\left| \begin{array}{l} 1\\ 0 \end{array} \right. + 4.\left. {\left( {\frac{{{x^3}}}{3} - \frac{{{x^2}}}{2}} \right)} \right|_{ - 1}^0 = \frac{7}{6} + \frac{{10}}{3} = \frac{9}{2}\)