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)

integral

1 - tan^{2} (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

\frac{x ^{3}}{( x + 2 ) ^{4}}

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\frac{x ^{4}}{( x ^{3} - 1 )}

\frac{1}{( 2 x ^{2} + x )}

\frac{sin ^{7} ( x )}{cos ^{2} ( x )}