Question
Evaluate the following intergrals:
$\int\text{e}^\text{ax}\sin(\text{bx}+\text{c})\text{dx}$
$\int\text{e}^\text{ax}\sin(\text{bx}+\text{c})\text{dx}$
✓
Answer
Let $\text{I}=\int\text{e}^\text{ax}\sin(\text{bx}+\text{c})\text{dx}$
$\Rightarrow-\text{e}^\text{ax}\frac{\cos(\text{bx}+\text{x})}{\text{b}}+\int\text{ae}^\text{ax}\frac{\cos(\text{bx}+\text{c})}{\text{b}}\text{dx}$
$=-\frac{1}{\text{b}}\text{e}^\text{ax}\cos(\text{bx}+\text{c})+\frac{\text{a}}{\text{b}}\int\text{e}^\text{ax}\cos(\text{bx}+\text{c})\text{dx}$
$=-\frac{1}{\text{b}}\text{e}^\text{ax}\cos(\text{bx}+\text{c})+\frac{\text{a}}{\text{b}}\Big[\int\text{e}^\text{ax}\frac{\sin(\text{bx}+\text{c})}{\text{b}}-\int\text{ae}^{\text{ax}}\frac{\sin(\text{bx}+\text{c})}{\text{b}}\text{dx}\Big]+\text{C}_1$
$=\frac{\text{e}^\text{ax}}{\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}\\-\frac{\text{a}^2}{\text{b}^2}\int\text{e}^\text{ax}\sin(\text{bx}+\text{c})\text{dx}+\text{C}_1$
$\Rightarrow\text{I}=\frac{\text{e}^\text{ax}}{\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}\\-\frac{\text{a}^2}{\text{b}^2}\text{I}+\text{C}_1$
$\Rightarrow\text{I}=\Big\{\frac{\text{a}^2+\text{b}^2}{\text{b}^2}\Big\}-\frac{\text{e}^\text{ax}}{\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}+\text{C}_1$
$\Rightarrow\text{I}=\frac{\text{e}^\text{ax}}{\text{a}^2+\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}+\text{C}_1$
$\Rightarrow-\text{e}^\text{ax}\frac{\cos(\text{bx}+\text{x})}{\text{b}}+\int\text{ae}^\text{ax}\frac{\cos(\text{bx}+\text{c})}{\text{b}}\text{dx}$
$=-\frac{1}{\text{b}}\text{e}^\text{ax}\cos(\text{bx}+\text{c})+\frac{\text{a}}{\text{b}}\int\text{e}^\text{ax}\cos(\text{bx}+\text{c})\text{dx}$
$=-\frac{1}{\text{b}}\text{e}^\text{ax}\cos(\text{bx}+\text{c})+\frac{\text{a}}{\text{b}}\Big[\int\text{e}^\text{ax}\frac{\sin(\text{bx}+\text{c})}{\text{b}}-\int\text{ae}^{\text{ax}}\frac{\sin(\text{bx}+\text{c})}{\text{b}}\text{dx}\Big]+\text{C}_1$
$=\frac{\text{e}^\text{ax}}{\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}\\-\frac{\text{a}^2}{\text{b}^2}\int\text{e}^\text{ax}\sin(\text{bx}+\text{c})\text{dx}+\text{C}_1$
$\Rightarrow\text{I}=\frac{\text{e}^\text{ax}}{\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}\\-\frac{\text{a}^2}{\text{b}^2}\text{I}+\text{C}_1$
$\Rightarrow\text{I}=\Big\{\frac{\text{a}^2+\text{b}^2}{\text{b}^2}\Big\}-\frac{\text{e}^\text{ax}}{\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}+\text{C}_1$
$\Rightarrow\text{I}=\frac{\text{e}^\text{ax}}{\text{a}^2+\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}+\text{C}_1$
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