What is the integral of tan^2(8x) ?

The integral of tan^2(8x) is -x+1/8 tan(8x)+C

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integral of tan^2(8x)

\int tan^{2} (8 x) d x

- x + \frac{1}{8} tan (8 x) + C

Solution steps

\int tan^{2} (8 x) d x

Rewrite using trig identities

= \int - 1 + sec^{2} (8 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} (8 x) d x

\int 1 d x = x

\int sec^{2} (8 x) d x = \frac{1}{8} tan (8 x)

= - x + \frac{1}{8} tan (8 x)

Add a constant to the solution

= - x + \frac{1}{8} tan (8 x) + C

y = \frac{e ^{x}}{x - 1}

\frac{d y}{d x} = \frac{( e ^{- y} + e ^{- 5 x - y} )}{e ^{x} y}

x \to 1 lim (- 9 + 3 x^{2})

\int t^{2} - t^{4} \frac{d}{1} d t

y^{^{'''}} + 2 y^{^{''}} + 2 y^{^{'}} = 0

What is the integral of tan^2(8x) ?
The integral of tan^2(8x) is -x+1/8 tan(8x)+C