What is the integral of tan^2(3x) ?

The integral of tan^2(3x) is -x+1/3 tan(3x)+C

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integral of tan^2(3x)

\int tan^{2} (3 x) d x

- x + \frac{1}{3} tan (3 x) + C

Solution steps

\int tan^{2} (3 x) d x

Rewrite using trig identities

= \int - 1 + sec^{2} (3 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 1 d x + \int sec^{2} (3 x) d x

\int 1 d x = x

\int sec^{2} (3 x) d x = \frac{1}{3} tan (3 x)

= - x + \frac{1}{3} tan (3 x)

Add a constant to the solution

= - x + \frac{1}{3} tan (3 x) + C

\frac{d}{d x} (sin (x^{2 x}))

x \to 5 lim (\frac{∣ x - 5 ∣}{5 - x})

\frac{d}{d x} ((x + 5)^{3})

(0, 0), (- 2, - 6)

y^{^{'}} - 4 y = 3 e^{5 t}

What is the integral of tan^2(3x) ?
The integral of tan^2(3x) is -x+1/3 tan(3x)+C