Primitive of Exponential of a x by Cosine of b x/Proof 1

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Theorem

$\ds \int e^{a x} \cos b x \rd x = \frac {e^{a x} \paren {a \cos b x + b \sin b x} } {a^2 + b^2} + C$


Proof

\(\ds \int e^{a x} \cos b x \rd x\) \(=\) \(\ds \frac {e^{a x} \cos b x} a + \frac b a \int e^{a x} \sin b x \rd x\) Primitive of $e^{a x} \cos b x$: Lemma
\(\ds \) \(=\) \(\ds \frac {e^{a x} \cos b x} a + \frac b a \paren {\frac {e^{a x} \sin b x} a - \frac b a \int e^{a x} \cos b x \rd x}\) Primitive of $e^{a x} \sin b x$: Lemma
\(\ds \) \(=\) \(\ds \frac {e^{a x} a \cos b x + e^{a x} b \sin b x} {a^2} - \frac {b^2} {a^2} \int e^{a x} \cos b x \rd x\) simplifying
\(\ds \leadsto \ \ \) \(\ds \paren {1 + \frac {b^2} {a^2} } \int e^{a x} \cos b x \rd x\) \(=\) \(\ds \frac {e^{a x} \paren {a \cos b x + b \sin b x} } {a^2}\) simplifying
\(\ds \leadsto \ \ \) \(\ds \frac {a^2 + b^2} {a^2} \int e^{a x} \cos b x \rd x\) \(=\) \(\ds \frac {e^{a x} \paren {a \cos b x + b \sin b x} } {a^2}\) common denominator
\(\ds \leadsto \ \ \) \(\ds \int e^{a x} \cos b x \rd x\) \(=\) \(\ds \frac {e^{a x} \paren {a \cos b x + b \sin b x} } {a^2 + b^2}\) multiplying by $\dfrac {a^2} {a^2 + b^2}$

$\blacksquare$