What is the integral of tan^2(4x) ?

The integral of tan^2(4x) is -x+1/4 tan(4x)+C

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integral of tan^2(4x)

\int tan^{2} (4 x) d x

- x + \frac{1}{4} tan (4 x) + C

Solution steps

\int tan^{2} (4 x) d x

Rewrite using trig identities

= \int - 1 + sec^{2} (4 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} (4 x) d x

\int 1 d x = x

\int sec^{2} (4 x) d x = \frac{1}{4} tan (4 x)

= - x + \frac{1}{4} tan (4 x)

Add a constant to the solution

= - x + \frac{1}{4} tan (4 x) + C

\int x x d x

\int_{- \infty}^{3} e^{0.0001 t} d t

\frac{\partial}{\partial y} (\frac{x}{y})

\frac{d}{d x} (e^{a x^{8}})

f (x) = - \frac{4 x}{x ^{2} + 1}, at x = - 1

What is the integral of tan^2(4x) ?
The integral of tan^2(4x) is -x+1/4 tan(4x)+C