AP Calculus BC

Integral of e^x cos(x): Answer, Proof, and Steps

Q: Why does integrating e^x cos(x) require parts twice?

Because one pass trades \cos x for \sin x but leaves an e^x times a trig function. A second pass brings back the original integral, which you then solve for algebraically.

Q: What is the integral of e^x sin(x)?

\int e^x\sin x\,dx = \frac{e^x(\sin x - \cos x)}{2} + C. The same double-parts method applies, and only the internal sign differs from the cosine case.

The integral of e^x cos(x) is e^x(sin x + cos x)/2 + C. Integration by parts applied twice brings back the original integral, and solving that equation for it produces the factor of one half. Differentiating e^x(sin x + cos x)/2 returns e^x cos(x).

\int e^x\cos x\,dx = \frac{e^x(\sin x+\cos x)}{2} + C

Integration by parts, applied twice

Neither factor of $e^x\cos x$ vanishes under differentiation, so a single integration by parts will not finish. Applying it twice loops the integral back to itself.

\int u\,dv = uv - \int v\,du

Call the target integral $I$ . First take $u = \cos x$ and $dv = e^x\,dx$ , so $du = -\sin x\,dx$ and $v = e^x$ .

I = \int e^x\cos x\,dx = e^x\cos x + \int e^x\sin x\,dx

Apply parts again to the new integral, keeping $dv = e^x\,dx$ . Take $u = \sin x$ , so $du = \cos x\,dx$ and $v = e^x$ .

\int e^x\sin x\,dx = e^x\sin x - \int e^x\cos x\,dx = e^x\sin x - I

Solving for the integral

The original integral $I$ has reappeared on the right. Substitute the second result into the first and collect the $I$ terms.

I = e^x\cos x + e^x\sin x - I

2I = e^x(\sin x + \cos x)

Divide by $2$ to isolate $I$ , then attach the constant of integration.

\int e^x\cos x\,dx = \frac{e^x(\sin x + \cos x)}{2} + C

Where the one half comes from

The integral is not integrated away; it is solved for like a variable. Moving $-I$ across the equation creates the $2I$ , and dividing gives the $\frac{1}{2}$ .

Common mistakes

This double-parts pattern trips students in a few predictable ways.

Switching the choice of $u$ on the second pass. Both integrations by parts must keep $dv = e^x\,dx$ ; swapping to $u = e^x$ unwinds the first step back to $I = I$ .
Forgetting to divide by $2$ . After $2I = e^x(\sin x + \cos x)$ you must halve both sides.
Dropping the constant. Even after solving algebraically, the indefinite integral needs $+ C$ .

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An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

Why does integrating e^x cos(x) require parts twice?

Because one pass trades $\cos x$ for $\sin x$ but leaves an $e^x$ times a trig function. A second pass brings back the original integral, which you then solve for algebraically.

What is the integral of e^x sin(x)?

$\int e^x\sin x\,dx = \frac{e^x(\sin x - \cos x)}{2} + C$ . The same double-parts method applies, and only the internal sign differs from the cosine case.

What is the integral of e^x cos(x) from 0 to pi?

$\int_0^{\pi} e^x\cos x\,dx = \left[\frac{e^x(\sin x + \cos x)}{2}\right]_0^{\pi} = \frac{e^{\pi}(0 - 1)}{2} - \frac{1(0 + 1)}{2} = -\frac{e^{\pi} + 1}{2}$ .