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Câu hỏi: Biết $F\left( x \right)=\int{{{\text{e}}^{x}}\cos x\text{d}x={{\text{e}}^{x}}\left( A\sin x+B\cos x \right)+C}$ với $A$, $B$, $C$ $\in \mathbb{R}$. Giá trị của $A+B$ bằng
A. $-2$.
B. $-1$.
C. $2$.
D. $1$.
Xét $\int{{{\text{e}}^{x}}\cos x\text{d}x}$. Đặt $\left\{ \begin{aligned}
& u={{\text{e}}^{x}} \\
& \text{d}v=\cos x\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u={{\text{e}}^{x}}\text{d}x \\
& v=\sin x \\
\end{aligned} \right.$.
Khi đó: $\int{{{\text{e}}^{x}}\cos x\text{d}x={{\text{e}}^{x}}.\sin x-\int{{{\text{e}}^{x}}\sin x\text{d}x}}$.
Xét $\int{{{\text{e}}^{x}}\sin x\text{d}x}$. Đặt $\left\{ \begin{aligned}
& u={{\text{e}}^{x}} \\
& \text{d}v=\sin x\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u={{\text{e}}^{x}}\text{d}x \\
& v=-\cos x \\
\end{aligned} \right.$.
Khi đó: $\int{{{\text{e}}^{x}}\sin x\text{d}x=-{{\text{e}}^{x}}\cos x+\int{{{\text{e}}^{x}}\cos x\text{d}x}}$.
Do đó: $\int{{{\text{e}}^{x}}\cos x\text{d}x={{\text{e}}^{x}}.\sin x-\left( -{{\text{e}}^{x}}\cos x+\int{{{\text{e}}^{x}}\cos x\text{d}x} \right)}$.
Vậy $F\left( x \right)=\int{{{\text{e}}^{x}}\cos x\text{d}x}={{\text{e}}^{x}}\left( \dfrac{1}{2}\sin x+\dfrac{1}{2}\cos x \right)+C$. Suy ra $A=B=\dfrac{1}{2}\Rightarrow A+B=1$.
A. $-2$.
B. $-1$.
C. $2$.
D. $1$.
Xét $\int{{{\text{e}}^{x}}\cos x\text{d}x}$. Đặt $\left\{ \begin{aligned}
& u={{\text{e}}^{x}} \\
& \text{d}v=\cos x\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u={{\text{e}}^{x}}\text{d}x \\
& v=\sin x \\
\end{aligned} \right.$.
Khi đó: $\int{{{\text{e}}^{x}}\cos x\text{d}x={{\text{e}}^{x}}.\sin x-\int{{{\text{e}}^{x}}\sin x\text{d}x}}$.
Xét $\int{{{\text{e}}^{x}}\sin x\text{d}x}$. Đặt $\left\{ \begin{aligned}
& u={{\text{e}}^{x}} \\
& \text{d}v=\sin x\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u={{\text{e}}^{x}}\text{d}x \\
& v=-\cos x \\
\end{aligned} \right.$.
Khi đó: $\int{{{\text{e}}^{x}}\sin x\text{d}x=-{{\text{e}}^{x}}\cos x+\int{{{\text{e}}^{x}}\cos x\text{d}x}}$.
Do đó: $\int{{{\text{e}}^{x}}\cos x\text{d}x={{\text{e}}^{x}}.\sin x-\left( -{{\text{e}}^{x}}\cos x+\int{{{\text{e}}^{x}}\cos x\text{d}x} \right)}$.
Vậy $F\left( x \right)=\int{{{\text{e}}^{x}}\cos x\text{d}x}={{\text{e}}^{x}}\left( \dfrac{1}{2}\sin x+\dfrac{1}{2}\cos x \right)+C$. Suy ra $A=B=\dfrac{1}{2}\Rightarrow A+B=1$.
Đáp án D.