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

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

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integral of 1-tan^2(x)

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

2 x - tan (x) + C

Solution steps

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

\int 1 d x = x

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

= x - (- x + tan (x))

Simplify

= 2 x - tan (x)

Add a constant to the solution

= 2 x - tan (x) + C

y^{^{''}} + 11 y^{^{'}} + 28 y = 0

\frac{d}{d x} ((3 x - 2)^{2})

x \to 0 lim (ab s x)

\frac{\partial}{\partial v} (u + v^{2})

\frac{d y}{d x} + y = - e^{x}