# Calculate Integral [(1+(tgx)^2)/tg x]dx and specify the method.

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We know that (tanx)' = (sec^2 x)dx = (1+tan^2 x) dx. Therefore,

Integral [(1+tan^2)dx/tanx] = integral d(tanx)'/tanx = ln(tanx), as Integral f'(x)dx/f(x) = ln[f(x)].

First of all, we'll notice that 1+(tgx)^2 = 1/(cos x)^2, from the fundamental formula of trigonometry:

(sin x)^2 + (cos x)^2 = 1

(sin x)^2/(cos x)^2 + 1 = 1/(cos x)^2

(tg x)^2 + 1 = 1/(cos x)^2

Int [(1+(tgx)^2)/tg x]dx=Int dx/(tg x)(cos x)^2

Now, we can choose the method of substitution.

tg x = t, so, differentiating, we'll have:

dx/(cos x)^2 = dt

**Int (1/t)dt = ln t + C = ln (tg x) + C**