Let #160;I= #8747; #160;eaxcosbxdxConsidering #160;cos #160;bx #160;as #160;first #160;function #160;and #160;eax #160;as #160;second ...

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Property 6

∫eaxcos b x d...

Question

$\int e^{a x} \cos b x d x$

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Solution

$Let I = \int e^{a x} \cos (b x) d x Considering \cos (b x) as first function and e^{a x} as second function I = \cos (b x) \frac{e^{a x}}{a} - \int - \sin (b x) \times b \times \frac{e^{a x}}{a} d x \Rightarrow I = \frac{e^{a x} \cos (b x)}{a} + \frac{b}{a} \int \sin (b x) e^{a x} d x \Rightarrow I = \frac{e^{a x}}{a} \cos (b x) + \frac{b}{a} \int e^{a x} \times \sin (b x) d x \Rightarrow I = \frac{e^{a x}}{a} \cos (b x) + \frac{b}{a} I_{1} . . . . . (1) where I_{1} = \int e^{a x} \sin (b x) d x Now, I_{1} = \int e^{a x} \sin (b x) d x Considering \sin (b x) as first function e^{a x} as second function I_{1} = \sin (b x) \frac{e^{a x}}{a} - \int \cos (b x) b \frac{e^{a x}}{a} d x \Rightarrow I_{1} = \frac{\sin (b x) e^{a x}}{a} - \frac{b}{a} \int e^{a x} \cos (b x) d x \Rightarrow I_{1} = \frac{e^{a x} \sin (b x)}{a} - \frac{b}{a} I . . . . . (2) From (1) and (2) I = \frac{e^{a x}}{a} \cos (b x) + \frac{b}{a} [\frac{e^{a x} \sin (b x)}{a} - \frac{b}{a} I] \Rightarrow I = \frac{e^{a x} \cos (b x)}{a} + \frac{b e^{a x} \sin (b x)}{a^{2}} - \frac{b^{2}}{a^{2}} I \Rightarrow I (1 + \frac{b^{2}}{a^{2}}) = e^{a x} [\frac{a \cos (b x) + b \sin (b x)}{a^{2}}] ∴ I = \frac{e^{a x} [a \cos b x + b \sin (b x)]}{(a^{2} + b^{2})} + C$

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