AP Calculus BC

Integral of x cos x: Answer, Parts, Mistakes

Q: What is the integral of x cos x?

It is \cos x + x\sin x + C, equivalently x\sin x + \cos x + C. Integration by parts with u = x and dv = \cos x\,dx gives x\sin x - \int \sin x\,dx = x\sin x + \cos x + C.

Q: Why does x cos x need integration by parts?

Because it is a product of two unrelated factors. Neither x nor \cos x is the derivative of the other, so substitution has nothing to cancel. Parts reverses the product rule and reduces x\cos x to the elementary integral of \sin x.

Q: What is the definite integral of x cos x from 0 to pi/2?

It equals \frac{\pi}{2} - 1. Evaluate \cos x + x\sin x: at x = \frac{\pi}{2} it is 0 + \frac{\pi}{2}, and at x = 0 it is 1, so the difference is \frac{\pi}{2} - 1.

The integral of x cos x with respect to x is cos x + x sin x + C. Integration by parts with u = x and dv = cos x dx gives x sin x minus the integral of sin x, so the answer is x sin x + cos x + C. Differentiating it with the product rule returns x cos x.

\int x\cos x\,dx = \cos x + x\sin x + C

How to integrate x cos x by parts

The integrand pairs an algebraic factor with a trig factor, and neither is the derivative of the other, so substitution stalls and integration by parts takes over. Parts is the reverse of the product rule.

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

Let $u$ be the factor that simplifies when differentiated. Here $x$ becomes $1$ , while $\cos x$ only cycles, so take $u = x$ and $dv = \cos x\,dx$ .

u = x, \quad du = dx, \quad dv = \cos x\,dx, \quad v = \sin x

Substitute. The $uv$ term is $x\sin x$ , and the new integral $\int v\,du = \int \sin x\,dx$ is elementary.

\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x + C

Written with the cosine first, that is $\cos x + x\sin x + C$ .

Check it by differentiating

Differentiate $\cos x + x\sin x$ : $-\sin x + \left[\sin x + x\cos x\right] = -\sin x + \sin x + x\cos x = x\cos x$ . The sine terms cancel and the integrand comes back.

Why u = x, plus a definite example

Parts pays off only when $\int v\,du$ is simpler. Taking $u = \cos x$ and $dv = x\,dx$ makes $v = \frac{x^2}{2}$ and raises the power, so the leftover integral is worse than the original. LIATE confirms the choice: the algebraic $x$ outranks the trig $\cos x$ , so $u = x$ .

\int_{0}^{\pi/2} x\cos x\,dx = \left[\cos x + x\sin x\right]_{0}^{\pi/2} = \left(0 + \frac{\pi}{2}\right) - (1 + 0) = \frac{\pi}{2} - 1

This is a Topic 6.11 integration by parts problem in Unit 6, which weights 15 to 20 percent of the exam. Its close relative $\int x\sin x\,dx = \sin x - x\cos x + C$ uses the same setup with the roles of sine and cosine swapped.

Common mistakes with the integral of x cos x

Multiplying antiderivatives. There is no product rule for integrals, so $\sin x \cdot \frac{x^2}{2}$ is not the answer.
Getting the sign of $v$ wrong. Antidifferentiating $\cos x$ gives $v = \sin x$ ; it is differentiating that flips the sign, not antidifferentiating.
Dropping the minus before $\int v\,du$ and writing $x\sin x - \cos x + C$ . Differentiate to check and the error shows up at once.
Choosing $u = \cos x$ . Legal but self-defeating, since $v = \frac{x^2}{2}$ makes the remaining integral harder.

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

What is the integral of x cos x?

It is $\cos x + x\sin x + C$ , equivalently $x\sin x + \cos x + C$ . Integration by parts with $u = x$ and $dv = \cos x\,dx$ gives $x\sin x - \int \sin x\,dx = x\sin x + \cos x + C$ .

Why does x cos x need integration by parts?

Because it is a product of two unrelated factors. Neither $x$ nor $\cos x$ is the derivative of the other, so substitution has nothing to cancel. Parts reverses the product rule and reduces $x\cos x$ to the elementary integral of $\sin x$ .

What is the definite integral of x cos x from 0 to pi/2?

It equals $\frac{\pi}{2} - 1$ . Evaluate $\cos x + x\sin x$ : at $x = \frac{\pi}{2}$ it is $0 + \frac{\pi}{2}$ , and at $x = 0$ it is $1$ , so the difference is $\frac{\pi}{2} - 1$ .