Evaluate the indefinite integral of tan(x) using substitution

Let $u = \cos(x)$, so $du = -\sin(x)\,dx$, which means $\sin(x)\,dx = -du$. Substituting into the integral gives $\int \frac{\sin(x)}{\cos(x)}\,dx = \int \frac{-du}{u} = -\int \frac{1}{u}\,du.$

Integrating gives $-\int \frac{1}{u}\,du = -\ln|u| + C$. Substituting back $u = \cos(x)$ yields the final answer $\int \tan(x)\,dx = -\ln|\cos(x)| + C.$

The substitution used was u equals cosine of x. This choice was made because the derivative of cosine is negative sine, which matched the sine term already present in the numerator. That allowed the entire integral to be rewritten purely in terms of u before integrating. The answer is cosine.