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Nguyên văn bởi Kratos  Bài 2. Tìm nguyên hàm: $\int{\dfrac{1+\sin x}{1+\cos x}e^xdx} $ |
$I=\int{\frac{1+\sin x}{1+\cos x}{{e}^{x}}dx=}\int{\frac{1+\sin x}{1+\cos x}d({{e}^{x}})={{e}^{x}}\frac{1+\sin x}{1+\cos x}-\int{{{e}^{x}}d\left( \frac{1+\sin x}{1+\cos x} \right)}} $
$={{e}^{x}}\frac{1+\sin x}{1+\cos x}-\int{{{e}^{x}}\frac{1+\cos x+\sin x}{{{(1+\cos x)}^{2}}}dx=}{{e}^{x}}\frac{1+\sin x}{1+\cos x}-\int{\frac{{{e}^{x}}dx}{1+\cos x}-}\int{\frac{{{e}^{x}}\sin xdx}{{{(1+\cos x)}^{2}}}} $
Đặt $K=\int{\frac{{{e}^{x}}}{1+\cos x}dx,L=\int{\frac{{{e}^{x}}\sin x}{{{(1+\cos x)}^{2}}}dx}} $
Xét $ L=\int{\frac{{{e}^{x}}\sin x}{{{(1+\cos x)}^{2}}}dx} $
Đặt $\left\{ \begin{align}
& u={{e}^{x}} \\
& dv=\frac{\sin x}{{{(1+\cos x)}^{2}}}dx \\
\end{align} \right.\Rightarrow \left\{ \begin{align}
& du={{e}^{x}}dx \\
& v=\int{\frac{-d(1+\cos x)}{{{(1+\cos x)}^{2}}}=\frac{1}{1+\cos x}} \\
\end{align} \right. $
$\Rightarrow L=\frac{{{e}^{x}}}{1+\cos x}-\int{\frac{{{e}^{x}}}{1+\cos x}dx=}\frac{{{e}^{x}}}{1+\cos x}-K $
Vậy $I={{e}^{x}}\frac{1+\sin x}{1+\cos x}-K-\left( \frac{{{e}^{x}}}{1+\cos x}-K \right)+C={{e}^{x}}\frac{1+\sin x}{1+\cos x}-\frac{{{e}^{x}}}{1+\cos x}+C $
[RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT]