Tính tích phân sau: I=∫(e^(x^2)−cos x)xdx

Question

Tính tích phân sau:

\(I=\int _{-1}^{0}\left(e^{x^{2} } -\cos x\right)x{\rm d}x .\)

Answer 1

\(I=\int _{-1}^{0}\left(e^{x^{2} } -\cos x\right)xdx \). 
\(\int _{-1}^{0}\left(e^{x^{2} } -\cos x\right)xdx =\int _{-1}^{0}e^{x^{2} } xdx -\int _{-1}^{0}x\cos xdx  \)
Ta có: 

\( I=\int _{-1}^{0}e^{x^{2} } xdx =\frac{1}{2} \int _{-1}^{0}e^{x^{2} } \left(x^{2} \right)^{{'} } dx =\frac{1}{2} e^{x^{2} } \left|\begin{array}{l} {0} \\ {-1} \end{array}\right. =\frac{1-e}{2} \)

\(J=\int _{-1}^{0}x\cos xdx \)

 \(Đặt \left\{\begin{array}{l} {u=x} \\ {dv=\cos xdx} \end{array}\right. \Rightarrow \left\{\begin{array}{l} {du=dx} \\ {v=\sin x} \end{array}\right. \)
\(J=x\sin x\left|\begin{array}{l} {0} \\ {-1} \end{array}\right. -\int _{-1}^{0}\sin xdx  \)
\(=\sin \left(-1\right)+\cos x\left|\begin{array}{l} {0} \\ {-1} \end{array}\right. =\sin \left(-1\right)+1-\cos \left(-1\right) \)
Vậy \(\int _{-1}^{0}\left(e^{x^{2} } -\cos x\right)xdx \)

\(=\frac{1-e}{2} -\left(\sin \left(-1\right)+1-\cos \left(-1\right)\right)=\sin 1+\cos 1-\frac{1+e}{2} \)