Lời giải:
Ta có:
\(F(x)=\int f(x)dx=\int e^x\cos xdx\)
Đặt \(\left\{\begin{matrix} u=e^x\\ dv=\cos xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=e^xdx\\ v=\int \cos xdx=\sin x\end{matrix}\right.\)
Do đó:
\(F(x)=\int e^x\cos xdx=e^x\sin x-\int \sin x.e^xdx+c\) (1)
Đặt \(\left\{\begin{matrix} u=e^x\\ dv=\sin xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=e^xdx\\ v=\int \sin xdx=-cos x\end{matrix}\right.\)
\(\Rightarrow \int \sin x.e^xdx=-\cos x.e^x+\int \cos x.e^xdx+c\) (2)
Từ (1)(2) suy ra:
\(F(x)=e^x.\sin x+\cos x.e^x-\int \cos x.e^xdx+c\)
\(\Leftrightarrow F(x)=e^x\sin x+e^x\cos x-F(x)+c\)
\(\Leftrightarrow F(x)=\frac{1}{2}e^x(\sin x+\cos x)+c\)
Do đó: \(a=b=\frac{1}{2}\)
