AP Calculus BC

Integral of x sin 2x: Answer, Proof, and Steps

Q: Where does the plus sign on sin 2x over 4 come from?

From two minus signs meeting. v = -\frac{\cos 2x}{2} is negative, and the parts formula subtracts \int v\,du, so the leftover integral enters positive.

Q: Can u-substitution do this instead?

No. Setting u = 2x clears the inside coefficient but leaves \frac{1}{4}\int u\sin u\,du, still a product of two different kinds of function. The stray x is not a multiple of the derivative of anything inside, so substitution alone cannot finish it.

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

Evaluate -\frac{x\cos 2x}{2} + \frac{\sin 2x}{4} at both ends. At x = \frac{\pi}{2}, \cos \pi = -1 and \sin \pi = 0, giving \frac{\pi}{4}; at x = 0 both terms vanish. The integral is \frac{\pi}{4}.

The integral of x sin(2x) is -(x cos(2x))/2 + sin(2x)/4 + C. Integration by parts with u = x and dv = sin(2x) dx gives v = -cos(2x)/2, and the leftover integral of cos(2x)/2 supplies the sin(2x)/4 term.

\int x\sin 2x\,dx = -\frac{x\cos 2x}{2}+\frac{\sin 2x}{4} + C

Setting up parts

A polynomial multiplied by a sine is the standard integration by parts case. Differentiating $x$ removes it entirely, and integrating $\sin 2x$ keeps the trig factor the same size.

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

Two separate things happen inside $v$ : the antiderivative of sine carries a minus sign, and the inside coefficient 2 divides.

u = x, \quad du = dx, \quad dv = \sin 2x\,dx, \quad v = -\frac{\cos 2x}{2}

Finishing and checking

Substituting into the formula, the minus sign in $v$ meets the minus sign in front of $\int v\,du$ , so the remaining integral enters with a plus.

\int x\sin 2x\,dx = -\frac{x\cos 2x}{2} + \int \frac{\cos 2x}{2}\,dx

\int \frac{\cos 2x}{2}\,dx = \frac{1}{2}\cdot\frac{\sin 2x}{2} = \frac{\sin 2x}{4}

\int x\sin 2x\,dx = -\frac{x\cos 2x}{2} + \frac{\sin 2x}{4} + C

Differentiate to confirm. The product rule on the first term gives $-\frac{\cos 2x}{2} + x\sin 2x$ , and the second term gives $\frac{2\cos 2x}{4} = \frac{\cos 2x}{2}$ . The cosine pieces cancel, leaving $x\sin 2x$ .

The mistake students make

Almost every wrong answer here is a sign, not a method. Writing $v = \frac{\cos 2x}{2}$ without the minus sign negates the whole answer, since that one sign rides through both terms, while forgetting the minus in front of $\int v\,du$ flips only the second term. Differentiating separates them: the first slip returns $-x\sin 2x$ , and the second returns $x\sin 2x - \cos 2x$ .

The other slip is dropping the $\frac{1}{2}$ in $v$ , which comes from the inside coefficient. Both terms in the final answer inherit that half, which is why the second one lands on a 4 in the denominator rather than a 2.

One derivative settles it

Differentiating your answer takes about fifteen seconds and catches every sign error. If the product rule does not return $x\sin 2x$ exactly, the mistake is upstream in $v$ .

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

Where does the plus sign on sin 2x over 4 come from?

From two minus signs meeting. $v = -\frac{\cos 2x}{2}$ is negative, and the parts formula subtracts $\int v\,du$ , so the leftover integral enters positive.

Can u-substitution do this instead?

No. Setting $u = 2x$ clears the inside coefficient but leaves $\frac{1}{4}\int u\sin u\,du$ , still a product of two different kinds of function. The stray $x$ is not a multiple of the derivative of anything inside, so substitution alone cannot finish it.

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

Evaluate $-\frac{x\cos 2x}{2} + \frac{\sin 2x}{4}$ at both ends. At $x = \frac{\pi}{2}$ , $\cos \pi = -1$ and $\sin \pi = 0$ , giving $\frac{\pi}{4}$ ; at $x = 0$ both terms vanish. The integral is $\frac{\pi}{4}$ .