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Basic Differentiation Rule

sin b x+C d x

Question

$\int e^{a x} \sin (b x + C) d x$

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Solution

$Let I = \int e^{a x} \sin (b x + C) d x Considering \sin (b x + C) as first function and e^{a x} as second function I = \sin (b x + C) \frac{e^{a x}}{a} - \int \cos (b x + C) b \frac{e^{a x}}{a} d x \Rightarrow I = \frac{e^{a x} \sin (b x + C)}{a} - \frac{b}{a} \int e^{a x} \cos (b x + C) d x \Rightarrow I = \frac{e^{a x} \sin (b x + C)}{a} - \frac{b}{a} I_{1} . . . (1) where I_{1} = \int e^{a x} \cos (b x + C) d x Now, I_{1} = \int e^{a x} \cos (b x + C) d x Consider \cos (b x + C) as first function e^{a x} as second funciton I_{1} = \cos (b x + C) \frac{e^{a x}}{a} - \int - \sin (b x + C) b \frac{e^{a x}}{a} d x \Rightarrow I_{1} = \frac{e^{a x} \cos (b x + C)}{a} + \frac{b}{a} \int e^{a x} \sin (b x + C) d x \Rightarrow I_{1} = \frac{e^{a x} \cos (b x + C)}{a} + \frac{b}{a} I . . . . . (2) From (1) & (2) I = \frac{e^{a x} \sin (b x + C)}{a} - \frac{b}{a} [\frac{e^{a x} \cos (b x + C)}{a} + \frac{b}{a} I] \Rightarrow I = \frac{e^{a x} \sin (b x + C)}{a} - \frac{b}{a^{2}} e^{a x} co s (b x + C) - \frac{b^{2}}{a^{2}} I \Rightarrow I (1 + \frac{b^{2}}{a^{2}}) = \frac{e^{a x} a \sin (b x + C) - b e^{a x} \cos (b x + C)}{a^{2}} + C_{1} \Rightarrow I = e^{a x} \frac{[a \sin (b x + C) - b \cos (b x + C)]}{a^{2} + b^{2}} + C_{1} Where C_{1} is integration constant$

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