Evaluate the integral: I = integral e^axsin bx dx

Evaluate the integral: $$ I = \int e^{ax}\sin bx\,dx. $$

Step 1 · First integration by parts

Let:

$$ u = \sin bx \quad \Rightarrow \quad du = b\cos bx\,dx, \qquad dv = e^{ax}\,dx \quad \Rightarrow \quad v = \frac{1}{a}e^{ax}. $$

Applying the formula:

$$ I = \frac{1}{a}e^{ax}\sin bx - \frac{b}{a}\int e^{ax}\cos bx\,dx. $$

Step 2 · Second integration by parts

For $\displaystyle\int e^{ax}\cos bx\,dx$, let $u=\cos bx$, $dv=e^{ax}dx$, so $du=-b\sin bx\,dx$, $v=\tfrac1ae^{ax}$:

$$ \int e^{ax}\cos bx\,dx = \frac{1}{a}e^{ax}\cos bx + \frac{b}{a}\int e^{ax}\sin bx\,dx = \frac{1}{a}e^{ax}\cos bx + \frac{b}{a}I. $$

Step 3 · Solve for the integral

Substituting back:

$$ I = \frac{1}{a}e^{ax}\sin bx - \frac{b}{a}\left(\frac{1}{a}e^{ax}\cos bx + \frac{b}{a}I\right) = \frac{1}{a}e^{ax}\sin bx - \frac{b}{a^2}e^{ax}\cos bx - \frac{b^2}{a^2}I. $$

Collecting the $I$ terms:

$$ \frac{a^2+b^2}{a^2}I = \frac{1}{a}e^{ax}\sin bx - \frac{b}{a^2}e^{ax}\cos bx. $$

Step 4 · Solve and simplify

$$ I = \frac{1}{a^2+b^2}\left(ae^{ax}\sin bx - be^{ax}\cos bx\right) + C. $$

Summary Table

Additional Notes

• The integrand is defined for all real $x$ (assuming $a,b$ are nonzero constants).
• Setting $a=-1$, $b=1$ recovers Problem 168's result exactly, confirming this general formula's consistency: $\tfrac{1}{1+1}(-e^{-x}\sin x-e^{-x}\cos x) = -\tfrac12e^{-x}(\sin x+\cos x)$.
• The result can be checked by differentiating $\tfrac{1}{a^2+b^2}(ae^{ax}\sin bx-be^{ax}\cos bx)$ and confirming it returns $e^{ax}\sin bx$.