AP Calculus AB and BC

Integral of sin 5x: Answer and the Divided Factor

The integral of sin 5x is negative cos 5x over 5, plus C. The minus comes from antidifferentiating sine and the division by 5 compensates for the inner coefficient, which the chain rule would otherwise reintroduce.

∫sin⁡5x dx=−cos⁡5x5+C\int \sin 5x\,dx = -\frac{\cos 5x}{5} + C

Why the 5 divides

∫sin⁡(kx)dx=−cos⁡(kx)k+C\int \sin(kx)dx = -\frac{\cos(kx)}{k} + C

Differentiate the answer to see it: the chain rule multiplies by 55, which cancels the 15\frac{1}{5} and returns sin⁡5x\sin 5x. Without the division you would get 5sin⁡5x5\sin 5x.

The definite version

∫0π/5sin⁡5x dx=[−cos⁡5x5]0π/5=15+15=25\int_{0}^{\pi/5}\sin 5x\,dx = \left[-\frac{\cos 5x}{5}\right]_{0}^{\pi/5} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}

That interval is exactly one half-period, so the result is the area of a single arch, scaled by the frequency.

Common mistakes

  • Multiplying by 55 instead of dividing.
  • Losing the minus sign. Sine antidifferentiates to NEGATIVE cosine.
  • Writing −cos⁡x5-\frac{\cos x}{5}, moving the coefficient out of the argument.

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 sin 5x?

It is −cos⁡5x5+C-\frac{\cos 5x}{5} + C.

Why divide rather than multiply?

Differentiating the answer reintroduces the 55 by the chain rule, so the 15\frac{1}{5} is there to cancel it.