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

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

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integral of tan^2(3x-2)

integral

tan^{2} (3 x - 2)

- x + \frac{1}{3} tan (3 x - 2) + C

Solution steps

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

Rewrite using trig identities

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

\int 1 d x = x

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

= - x + \frac{1}{3} tan (3 x - 2)

Add a constant to the solution

= - x + \frac{1}{3} tan (3 x - 2) + C

in t e g r a l \frac{ln ( x )}{( 1 + ln ( x ) ) ^{2}}

in t e g r a l (1 + x)^{n}

in t e g r a l \frac{1}{1 - t ^{2}}

in t e g r a l a^{12} y^{4} - a^{4} y^{12}

in t e g r a l \frac{1}{( x + 4 ) ^{2}}

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