- \(\displaystyle \int \frac {\cos(x)}{\sin ^2 (x)} ~dx = \color {blue} {- \csc (x)}\)
- Solution: \(\frac {\cos (x)}{\sin ^2 (x)} = \cot (x) \csc (x)\)
Solution
1. The substitution \(u = \sin(x)\) gives \(\int \frac{\cos(x)}{\sin^2(x)} \, dx = -\csc(x) + C\).
With \(u = \sin(x)\), \(du = \cos(x) \, dx\), so \[\int \frac{\cos(x)}{\sin^2(x)} \, dx = \int u^{-2} \, du = -\frac{1}{u} + C = -\frac{1}{\sin(x)} + C = -\csc(x) + C.\]
2. The identity \(\frac{\cos(x)}{\sin^2(x)} = \cot(x)\csc(x)\) gives the same antiderivative.
\(\frac{\cos(x)}{\sin^2(x)} = \frac{\cos(x)}{\sin(x)} \cdot \frac{1}{\sin(x)} = \cot(x)\csc(x)\), and since \(\frac{d}{dx}(\csc(x)) = -\csc(x)\cot(x)\), \[\int \csc(x)\cot(x) \, dx = -\csc(x) + C.\]
3. Q.E.D.
By 1, \(\int \frac{\cos(x)}{\sin^2(x)} \, dx = \boxed{-\csc(x) + C}\); 2 checks it.