AP Calculus AB and BC

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

Q: Why is there no fraction in front of the answer?

Because du = \cos x\,dx matches the integrand exactly, with no constant left over. Coefficients like \frac{1}{4} appear only when du differs from what is present by a constant factor, as in \int e^{4x}\,dx.

Q: What happens without the cos x factor?

\int e^{\sin x}\,dx has no antiderivative in elementary functions. The \cos x is what makes this integral solvable, which is why AP problems always supply it.

The integral of e^(sin x)cos(x) is e^(sin x) + C. Letting u = sin x gives du = cos x dx, so the integral becomes the integral of e^u du, which is e^u. Differentiating e^(sin x) by the chain rule gives e^(sin x) times cos x, the original integrand.

\int e^{\sin x}\cos x\,dx = e^{\sin x} + C

Substituting u = sin x

The exponent is $\sin x$ and the factor outside is $\cos x$ , exactly the derivative of that exponent. Nothing needs to be adjusted by a constant.

u = \sin x, \qquad du = \cos x\,dx

\int e^{\sin x}\cos x\,dx = \int e^{u}\,du = e^{u} + C = e^{\sin x} + C

This is the pattern $\int e^{g(x)}g'(x)\,dx = e^{g(x)} + C$ . Whenever the exponent's derivative multiplies the exponential, the antiderivative is the exponential itself.

Checking by differentiating

Differentiate $e^{\sin x}$ with the chain rule: keep the exponential, then multiply by the derivative of the exponent.

\frac{d}{dx}e^{\sin x} = e^{\sin x}\cdot\cos x

That is the integrand, with no leftover constant, which is why the answer carries no coefficient in front.

The mistakes students make

Adding a coefficient such as $\frac{e^{\sin x}}{\cos x}$ . Dividing by an inner derivative is only valid when that derivative is a constant, and $\cos x$ is not.
Answering $e^{\sin x}\sin x$ by trying to antidifferentiate each factor. The $\cos x$ is consumed by $du$ , not integrated on its own.
Reaching for integration by parts. The integrand is already in substitution form, so parts only lengthens the work.
Answering $e^{\cos x}$ from a sign slip. Check by differentiating: $\frac{d}{dx}e^{\cos x} = -e^{\cos x}\sin x$ , which is not the integrand.

Every answer on this page is machine checked

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 is there no fraction in front of the answer?

Because $du = \cos x\,dx$ matches the integrand exactly, with no constant left over. Coefficients like $\frac{1}{4}$ appear only when $du$ differs from what is present by a constant factor, as in $\int e^{4x}\,dx$ .

What is the integral of e^(sin x)cos(x) from 0 to pi/2?

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

What happens without the cos x factor?

$\int e^{\sin x}\,dx$ has no antiderivative in elementary functions. The $\cos x$ is what makes this integral solvable, which is why AP problems always supply it.