AP Calculus BC

Integral of x sin x: Answer, Parts, Mistakes

Q: What is the integral of x sin x?

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

Q: What is the definite integral of x sin x from 0 to pi?

It equals \pi. Evaluate \sin x - x\cos x at the endpoints: at x = \pi it is 0 - \pi(-1) = \pi, and at x = 0 it is 0, so the difference is \pi.

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

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

How to integrate x sin x by parts

The integrand is an algebraic factor times a trig factor, and neither is the derivative of the other, so substitution has nothing to grab. That is the signal for integration by parts, which reverses the product rule.

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

Choose $u$ to be the factor that simplifies when differentiated. Here $x$ differentiates to $1$ , while $\sin x$ only cycles, so let $u = x$ and $dv = \sin x\,dx$ .

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

Substitute into the formula. The $uv$ term is $-x\cos x$ , and the new integral $\int v\,du$ is $\int -\cos x\,dx$ , which is elementary.

\int x\sin x\,dx = -x\cos x - \int (-\cos x)\,dx = -x\cos x + \int \cos x\,dx

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

Check it by differentiating

Differentiate $\sin x - x\cos x$ , using the product rule on the second term: $\cos x - \left[\cos x + x(-\sin x)\right] = \cos x - \cos x + x\sin x = x\sin x$ . The cosine terms cancel and you recover the integrand.

Why u = x and not u = sin x

Parts only helps if $\int v\,du$ is easier than the integral you started with. Taking $u = \sin x$ and $dv = x\,dx$ is legal but pushes the power up: $v = \frac{x^2}{2}$ , and the new integral $\int \frac{x^2}{2}\cos x\,dx$ is harder than the original.

The LIATE ordering encodes the right choice: pick $u$ as the first type present, going Logarithmic, Inverse trig, Algebraic, Trig, Exponential. In $x\sin x$ the algebraic $x$ outranks the trig $\sin x$ , so $u = x$ .

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

Integration by parts is Topic 6.11 inside Unit 6, which weights 15 to 20 percent of the exam. The definite form above is the common way this integrand appears, and the antiderivative keeps the endpoint arithmetic short.

Common mistakes with the integral of x sin x

Multiplying antiderivatives. There is no product rule for integrals, so $-\cos x \cdot \frac{x^2}{2}$ is not the answer.
Getting the sign of $v$ wrong. Antidifferentiating $\sin x$ gives $v = -\cos x$ , not $+\cos x$ .
Dropping the minus in front of $\int v\,du$ and writing $-x\cos x - \sin x + C$ . One differentiation catches it.
Choosing $u = \sin x$ . It is legal but makes the remaining integral harder, because $v = \frac{x^2}{2}$ raises the power.

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 sin x?

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

Why can't I use u-substitution for x sin x?

Substitution needs an inner function whose derivative is already in the integrand. In $x\sin x\,dx$ , neither $x$ nor $\sin x$ is the derivative of the other up to a constant, so there is nothing to substitute. A product of two unrelated factors calls for integration by parts.

What is the definite integral of x sin x from 0 to pi?

It equals $\pi$ . Evaluate $\sin x - x\cos x$ at the endpoints: at $x = \pi$ it is $0 - \pi(-1) = \pi$ , and at $x = 0$ it is $0$ , so the difference is $\pi$ .