What is the integral of sec^2(x)-8 ?

The integral of sec^2(x)-8 is tan(x)-8x+C

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integral of sec^2(x)-8

\int sec^{2} (x) - 8 d x

tan (x) - 8 x + C

Solution steps

\int sec^{2} (x) - 8 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 8 d x

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

\int 8 d x = 8 x

= tan (x) - 8 x

Add a constant to the solution

= tan (x) - 8 x + C

t a y l or (x - π)^{3}

n = 1 \sum \infty 3 n

\int x csc (x) cot (x) d x

\int_{0}^{8} x + 1 d x

\int sec (2 x + 1) tan (2 x + 1) d x