∫tan(x)dx
ln|secx|+c
We can solve this integral using substitution. Let u = cos(x), then du/dx = -sin(x) and dx = du/(-sin(x)). We can also express tan(x) in terms of u:
tan(x) = sin(x)/cos(x) = ((1 – cos^2(x))/cos(x))^(1/2) / cos(x)
= (1/cos(x) – cos(x)/cos(x))^(1/2)
= (1 – u^2)^(1/2) / u
Substituting into the original integral, we get:
∫tan(x)dx = ∫(1 – u^2)^(1/2) / u * (-du/sin(x))
= -∫(1 – u^2)^(-1/2) du
Using the formula ∫(1 – x^2)^(-1/2) dx = arcsin(x) + C, we can solve the above integral:
∫tan(x)dx = -arcsin(u) + C
= -arcsin(cos(x)) + C
Therefore, the final answer is:
∫tan(x)dx = -arcsin(cos(x)) + C, where C is the constant of integration.
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