What is the integral of 1tan^2(x) ?

The integral of 1tan^2(x) is -x+tan(x)+C

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integral of 1tan^2(x)

\int 1 tan^{2} (x) d x

- x + tan (x) + C

Solution steps

\int 1 \cdot tan^{2} (x) d x

Simplify

= \int tan^{2} (x) d x

Rewrite using trig identities

= \int - 1 + sec^{2} (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int 1 d x + \int sec^{2} (x) d x

\int 1 d x = x

\int sec^{2} (x) d x = tan (x)

= - x + tan (x)

Add a constant to the solution

= - x + tan (x) + C

g (x) = 14 (1. 7^{x})

\frac{\partial}{\partial y} (e^{3 x})

x \to 2 - lim (\frac{x}{∣ x ∣})

\int \frac{x ^{2} - 1}{x ^{2} - 4} d x

x \to \infty lim (\frac{3 e ^{x}}{9 + 2 e ^{5 x}})