What is the integral of (sec^2(x)-9) ?

The integral of (sec^2(x)-9) is tan(x)-9x+C

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integral of (sec^2(x)-9)

\int (sec^{2} (x) - 9) d x

tan (x) - 9 x + C

Solution steps

\int sec^{2} (x) - 9 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 sec^{2} (x) d x - \int 9 d x

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

\int 9 d x = 9 x

= tan (x) - 9 x

Add a constant to the solution

= tan (x) - 9 x + C

\int_{0}^{2} \frac{3 x ^{2} - 1}{x ^{2}} d x

y = 2 x^{3} - x^{2} + 7

\frac{\partial}{\partial x} (\frac{t}{a ^{2} t ^{2} - x ^{2}})

\frac{e ^{x} + 1}{e ^{x} - 1}

\frac{d y}{d x} = 2 x y^{2} + 128 x

What is the integral of (sec^2(x)-9) ?
The integral of (sec^2(x)-9) is tan(x)-9x+C