Evaluate the following intergrals: ^ax dx - Vidyadip

Question

Evaluate the following intergrals:
$\int\text{e}^\text{ax}\cos\text{bx dx}$

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Answer

Let $\text{I}=\int\text{e}^\text{ax}\cos\text{bx dx}$
Intergrating by parts,
$\text{I}=\text{e}^\text{ax}\frac{\sin\text{bx}}{\text{b}}-\text{a}\int\text{e}^\text{ax}\frac{\sin\text{bx}}{\text{x}}\text{dx}$
$=\frac{1}{\text{b}}\text{e}^\text{ax}\sin\text{bx}-\frac{\text{a}}{\text{b}}\int\text{e}^\text{ax}\sin\text{bx dx}$
$=\frac{1}{\text{a}}\text{e}^\text{ax}\sin\text{bx}-\frac{\text{a}}{\text{b}}\Big[-\text{e}^\text{ax}\frac{\cos\text{bx}}{\text{b}}+\int\text{ae}^\text{ax}\frac{\cos\text{bx}}{\text{b}}\text{dx}\Big]$
$=\frac{1}{\text{a}}\text{e}^\text{ax}\sin\text{bx}-\frac{\text{a}}{\text{b}^2}\text{e}^\text{ax}\cos\text{bx}-\frac{\text{a}^2}{\text{b}^2}\int\text{e}^\text{ax}\cos\text{bx dx}$
$\Rightarrow\text{I}=\frac{\text{e}^\text{ax}}{\text{b}^2}\big[\text{b}\sin\text{bx}+\text{a}\cos\text{bx}\big]-\frac{\text{a}^2}{\text{b}^2}\text{I}+\text{C}$
$\Rightarrow\text{I}.\Big\{\frac{\text{a}^2+\text{b}^2}{\text{b}^2}\Big\}=\frac{\text{e}^\text{ax}}{\text{b}^2}\big[\text{b}\cos\text{bx}+\text{a}\cos\text{bx}\big]+\text{C}$
Thus,
$\text{I}=\frac{\text{e}^\text{ax}}{\text{a}^2+\text{b}^2}\big[\text{a}\cos\text{bx}+\text{a}\cos\text{bx}\big]+\text{C}$

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