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

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

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

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

- 9 sec (x) - 2 tan (x) + C

Solution steps

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

\int 9 sec (x) tan (x) d x = 9 sec (x)

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

= - 9 sec (x) - 2 tan (x)

Add a constant to the solution

= - 9 sec (x) - 2 tan (x) + C

\frac{\partial}{\partial x} (e^{x} - e^{- x})

\frac{d}{d x} (1 - \frac{1}{2 x})

\frac{d y}{d x} = \frac{x - 1}{y}

\int \frac{x ^{2}}{x ^{2} - 16} d x

\int \frac{1}{( a + b x ) ^{3}} d x

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