Integral of sin 2x / sin x: Simplify First

Identify before you integrate

No integration technique attacks $\frac{\sin 2x}{\sin x}$ directly, and none needs to, because this is not a problem about technique. The double angle identity $\sin 2x = 2\sin x\cos x$ removes the quotient entirely.

The holes that cancelling removes

The original integrand is undefined wherever $\sin x = 0$, that is at $x = 0$, $\pm\pi$, $\pm 2\pi$ and so on, while $2\cos x$ is defined everywhere. Cancelling patches a set of removable discontinuities.

For a definite integral this matters. $\int_{1}^{2}\frac{\sin 2x}{\sin x}dx = 2\sin 2 - 2\sin 1$ is fine, because the interval avoids the trouble spots. An interval $[a,b]$ containing $x = \pi$ does not, so it has to be split at the hole and evaluated with limits. Because the discontinuity is removable and the integrand stays bounded there, the value is still $2\sin b - 2\sin a$.

The mistakes students make

All three of these happen before a single integral sign is touched.

• Cancelling the sines as though $\frac{\sin 2x}{\sin x} = 2$, which gives the answer $2x + C$.
• Simplifying correctly to $2\cos x$ and then integrating with the derivative rule, producing $-2\sin x + C$. Cosine integrates to positive sine.
• Reading $\sin 2x$ as $\sin^{2}x$. The quotient then collapses to $\sin x$ rather than $2\cos x$, and the answer comes out $-\cos x + C$ instead of $2\sin x + C$.