AP Calculus AB and BC

Integral of cot 2x: Answer and Substitution

The integral of cot 2x is one half times the natural log of the absolute value of sin 2x, plus C. Substituting u equals sin 2x gives du equal to 2 cos 2x dx, and unlike the tangent case no minus sign appears.

∫cot⁡2x dx=12ln⁡∣sin⁡2x∣+C\int \cot 2x\,dx = \frac{1}{2}\ln\left|\sin 2x\right| + C

Substituting the denominator

u=sin⁡2x,du=2cos⁡2x dx  ⟹  ∫cos⁡2xsin⁡2x dx=12ln⁡∣sin⁡2x∣+Cu = \sin 2x, \quad du = 2\cos 2x\,dx \implies \int \frac{\cos 2x}{\sin 2x}\,dx = \frac{1}{2}\ln\left|\sin 2x\right| + C

Why no minus here

For tangent the substitution is on cosine, whose derivative carries a minus. For cotangent it is on sine, whose derivative does not. That single difference is the whole reason the two answers differ in sign.

Domain

cot⁡2x\cot 2x is undefined wherever sin⁡2x=0\sin 2x = 0, so the antiderivative is valid only on the intervals between those points, and a definite integral cannot cross one.

Common mistakes

  • Writing a minus sign by analogy with tangent. Only the tangent case has one.
  • Losing the 12\frac{1}{2} from the inner coefficient.
  • Substituting u=cos⁡2xu = \cos 2x instead, which does not match the numerator.

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 cot 2x?

It is 12ln⁡∣sin⁡2x∣+C\frac{1}{2}\ln\left|\sin 2x\right| + C.

Why is there no minus sign?

Because the substitution is on sin⁡2x\sin 2x, whose derivative is positive. The tangent case substitutes cosine, which brings a minus.