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

\int_{- 1}^{2} \frac{1}{x ^{4}} d x

y^{^{''}} + 2 y^{^{'}} + y = 0, y (1) = 1, y^{^{'}} (1) = 0

\frac{d}{d x} (x^{4} - 4 x^{3} + 3 x^{2} - 3)

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

x \to - 2 lim (7 x^{2} + 8)