What is the integral of (sec^2(x)-5) ?

The integral of (sec^2(x)-5) is tan(x)-5x+C

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integral of (sec^2(x)-5)

\int (sec^{2} (x) - 5) d x

tan (x) - 5 x + C

Solution steps

\int sec^{2} (x) - 5 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 5 d x

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

\int 5 d x = 5 x

= tan (x) - 5 x

Add a constant to the solution

= tan (x) - 5 x + C

f (x) = (1 + 5 x)^{9}, at x = 0

x \to 0 + lim (\frac{I n x}{x})

\frac{d y}{d x} = \frac{( x y + 3 x - y - 3 )}{x y - 2 x + 4 y - 8}

\int \frac{2 x}{x ^{2} + x} d x

\frac{s}{s + 4}

What is the integral of (sec^2(x)-5) ?
The integral of (sec^2(x)-5) is tan(x)-5x+C