Integrals of the Form ∫e^{ax}sin(bx)dx and ∫e^{ax}cos(bx)dx

A genuinely different situation arises with integrands like $e^{ax}\sin bx$ or $e^{ax}\cos bx$: applying integration by parts does not make the integral simpler or terminate the way it does for $x^ne^{ax}$ — instead, after two applications, the original integral reappears on the right-hand side, and the trick is to solve for it algebraically rather than keep integrating forever.

Result 11.1.

Proof (own words), part (i). Let $I=\displaystyle\int e^{ax}\sin bx\,dx$. Take $u=\sin bx$ ($du=b\cos bx\,dx$) and $dv=e^{ax}dx$ ($v=\dfrac{e^{ax}}a$); integration by parts gives

Now apply integration by parts a second time to the remaining integral, this time with $u=\cos bx$ ($du=-b\sin bx\,dx$) and the same $dv=e^{ax}dx$:

Substituting this back into the expression for $I$:

The unknown $I$ now appears on both sides, exactly the payoff of doing two by-parts steps with the same pairing convention throughout (per the Caution of §11.7.6). Collecting the $I$ terms,

and multiplying both sides by $\dfrac{a^2}{a^2+b^2}$ gives $I=\dfrac{e^{ax}}{a^2+b^2}[a\sin bx-b\cos bx]+c$, as stated. The cosine version follows by the identical two-step argument with the roles of sine and cosine exchanged at the first step.

Using the result. With $a$ and $b$ read off the given integral, the two boxed formulas are applied directly — e.g. for $\int e^{3x}\cos2x\,dx$, take $a=3,b=2$ to get $\dfrac{e^{3x}}{13}(3\cos2x+2\sin2x)+c$; for $\int e^{-5x}\sin3x\,dx$, take $a=-5,b=3$ to get $-\dfrac{e^{-5x}}{34}(5\sin3x+3\cos3x)+c$.

Result 11.2. A second, unrelated shortcut is useful whenever the integrand has the specific shape $e^x$ times a sum of a function and its own derivative:

Proof (own words). Split the integral into two pieces, $\displaystyle\int e^xf(x)\,dx+\int e^xf'(x)\,dx$. Apply integration by parts to the first piece only, with $u=f(x)$ ($du=f'(x)dx$) and $dv=e^xdx$ ($v=e^x$): $\displaystyle\int e^xf(x)\,dx=e^xf(x)-\int e^xf'(x)\,dx$. Substituting this back,

since the two copies of $\int e^xf'(x)dx$ exactly cancel. …