What is the integral of tan^2(ax) ?

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

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integral of tan^2(ax)

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

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

Solution steps

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

Rewrite using trig identities

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

\int 1 d x = x

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

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

Add a constant to the solution

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

\int \frac{( 1 - 2 x )}{x - x ^{2}} d x

\int \frac{x ^{2} - 3 x + 5}{x ^{3}} d x

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

\int \frac{x ^{4} + x ^{3} - x ^{2} - x + 1}{x ^{3} - x} d x

\int (\frac{32 x ^{3}}{2 x ^{4} + 5}) d x

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