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

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

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integral of tan^2(3-2x)

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{1}{2} tan^{2} (u) d u

= \int - \frac{1}{2} tan^{2} (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= - \frac{1}{2} \cdot \int - 1 + sec^{2} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \frac{1}{2} (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

\int sec^{2} (u) d u = tan (u)

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

Substitute back

u = 3 - 2 x

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

Simplify

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

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

Add a constant to the solution

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

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