AP Calculus AB and BC

Integral of sin x/(1+cos x)

The integral of sin x over 1 plus cos x is negative the natural log of the absolute value of 1 plus cos x, plus C. The denominator differentiates to negative sin x, so the substitution works with a minus sign out front.

∫sin⁡x1+cos⁡x dx=−ln⁡∣1+cos⁡x∣+C\int \frac{\sin x}{1+\cos x}\,dx = -\ln\left|1+\cos x\right| + C

Substituting the denominator

u=1+cos⁡x,du=−sin⁡x dx  ⟹  ∫sin⁡x1+cos⁡xdx=−ln⁡∣1+cos⁡x∣+Cu = 1+\cos x, \quad du = -\sin x\,dx \implies \int \frac{\sin x}{1+\cos x}dx = -\ln\left|1+\cos x\right| + C

The minus is not optional and it is not a convention: it is the derivative of cosine showing up in dudu.

The denominator can vanish

1+cos⁡x1+\cos x is zero at odd multiples of π\pi, where the integrand has a vertical asymptote. The antiderivative is valid only between those points.

Common mistakes

  • Losing the minus sign.
  • Substituting u=cos⁡xu = \cos x and then forgetting the 1+1+ when writing the answer.
  • Splitting the fraction into separate integrals.

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 x/(1+cos x)?

It is −ln⁡∣1+cos⁡x∣+C-\ln\left|1+\cos x\right| + C.

Why the minus sign?

Because ddx(1+cos⁡x)=−sin⁡x\frac{d}{dx}(1+\cos x) = -\sin x, so the substitution introduces it.