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

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

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integral of-9tan^2(x)

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Rewrite using trig identities

= - 9 \cdot \int - 1 + 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

= - 9 (- \int 1 d x + \int sec^{2} (x) d x)

\int 1 d x = x

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

= - 9 (- x + tan (x))

Add a constant to the solution

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

f (x) = x, at x = 16

\frac{d}{d x} (\frac{1}{1 + x ^{12}})

\frac{d}{d x} (1 - e^{- 8 x})

2 x + 5 x^{- 1}

\int_{1}^{4} (4 x^{2} - 2 x) d x

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