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Prove integration formulas for eaxcos (bx) and eaxsin (bx)

Let a and b be constants with at least one of them nonzero and define

    \[ A = \int e^{ax} \cos (bx) \, dx, \qquad B = \int e^{ax} \sin (bx) \, dx. \]

Using integration by parts, establish the following formulas for constants C_1, C_2,

    \[ aA - bB = e^{ax} \cos (bx) + C_1, \qquad aB + bA = e^{ax} \sin (bx) + C_2. \]

Using these formulas prove the following integration formulas,

    \begin{align*}  \int e^{ax} \cos (bx) \, dx &= \frac{e^{ax} (a \cos (bx) + b \sin (bx))}{a^2 + b^2}  + C, \\  \int e^{ax} \sin (bx) \, dx &= \frac{e^{ax} (a \sin (bx) - b \cos (bx))}{a^2 + b^2} + C. \end{align*}


To establish the formula aA - bB = e^{ax} \cos (bx) + C_1 we use integration by parts letting

    \begin{align*}  u &= \cos (bx), & du &= -b \sin (bx) \, dx \\  dv &= e^{ax} \, dx & v &= \frac{1}{a} e^{ax}. \end{align*}

Then we can evaluate A using the formula for integration by parts,

    \begin{align*}  &&A &= \int e^{ax} \cos (bx) \, dx \\[9pt]  \implies && A&= \frac{1}{a}e^{ax} \cos (bx) + \frac{b}{a} \int e^{ax} \sin (bx) \, dx + C_1 \\[9pt]  \implies && aA &= e^{ax} \cos (bx) + bB + C_1 \\[9pt]  \implies && aA - bB &= e^{ax} \cos (bx) + C_1. \end{align*}

To establish the second formula, aB + bA = e^{ax} \sin (bx) + C_2, we use integration by parts again. Let

    \begin{align*}  u &= \sin (bx) & du &= b \cos (bx) \, dx \\ dv &= e^{ax} \, dx & v &= \frac{1}{a} e^{ax}. \end{align*}

Then we have

    \begin{align*}  && B&= \int e^{ax} \sin (bx) \, dx \\[9pt] \implies && B &= \frac{1}{a} e^{ax} \sin (bx) - \frac{b}{a} \int e^{ax} \cos (bx) \, dx + C_2 \\[9pt]  \implies && aB &= e^{ax} \sin (bx) - bA + C_2 \\[9pt]  \implies && aB + bA &= e^{ax} \sin (bx) + C_2. \end{align*}

This establishes the two requested equations, now we prove the two integral identities.

Proof. Solving for B in the second equation above we have

    \[ aB + bA = e^{ax} \sin (bx) + C_2 \quad \implies \quad B = \frac{1}{a} \left( e^{ax} \sin (bx) - bA) + C_2.  \]

Plugging this into the first equation we have

    \begin{align*}  && aA - bB &= e^{ax} \cos (bx) + C_1 \\[9pt] \implies && aA - \frac{b}{a} \left( e^{ax} \sin (bx) - bA + C_2 \right) &= e^{ax} \cos (bx) + C_1 \\[9pt] \implies && aA - \frac{b}{a} e^{ax} \sin (bx) + \frac{b^2}{a} A &= e^{ax} \cos (bx) + C \\[9pt] \implies && A \left( a + \frac{b^2}{a}\right) &= e^{ax} \cos (bx) + \frac{b}{a} e^{ax} \sin (bx) + C \\[9pt] \implies && A (a^2 + b^2) &= a e^{ax} \cos (bx) + b e^{ax} \sin (bx) + C \\[9pt] \implies && A &= \frac{a e^{ax} \cos (bx) + b e^{ax} \sin (bx)}{a^2+b^2} + C \\[9pt] \implies && \int e^{ax} \cos (bx) \, dx &= \frac{a e^{ax} \cos (bx) + b e^{ax} \sin (bx)}{a^2+b^2} + C. \end{align*}

Next, for the second integral equation we are asked to prove, we use the formula we obtained for B above,

    \[ B = \frac{1}{a} \left( e^{ax} \sin (bx) - bA \right) + C. \]

Then, we use the expression we obtained for A into this,

    \[ B = \frac{1}{a} \left( e^{ax} \sin (bx) - b \left( \frac{a e^{ax} \cos (bx) + b e^{ax} \sin (bx)}{a^2+b^2} \right) \right) + C. \]

This implies,

    \begin{align*}  && B &= \frac{1}{a} e^{ax} \sin (bx) - \frac{be^{ax} \cos (bx) + \frac{b^2}{a} e^{ax} \sin (bx)}{a^2+b^2} + C \\[9pt] \implies && B &= \frac{(a^2+b^2)e^{ax} \sin (bx) - (ab)e^{ax} \cos (bx) - b^2 e^{ax} \sin (bx)}{a(a^2+b^2)} + C \\[9pt] \implies && B &= \frac{a^2 e^{ax} \sin (bx) - (ab) e^{ax} \cos (bx)}{a(a^2+b^2)} + C \\[9pt] \implies && B &= \frac{a e^{ax} \sin (bx) - b e^{ax} \cos (bx)}{a^2+b^2} + C \\[9pt] \implies && \int e^{ax} \sin (bx) \, dx &= \frac{a e^{ax} \sin (bx) - b e^{ax} \cos (bx)}{a^2+b^2} + C. \qquad \blacksquare \end{align*}

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