What is the integral of tan^2(3x-1)sec^2(3x-1) ?

The integral of tan^2(3x-1)sec^2(3x-1) is 1/9 tan^3(3x-1)+C

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integral of tan^2(3x-1)sec^2(3x-1)

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

\frac{1}{9} tan^{3} (3 x - 1) + C

Solution steps

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

Apply u-substitution

: \int tan^{2} (u) sec^{2} (u) \frac{1}{3} d u

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

Take the constant out:

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

= \frac{1}{3} \cdot \int tan^{2} (u) sec^{2} (u) d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int (1 + tan^{2} (u)) tan^{2} (u) d u

Apply u-substitution

: \int v^{2} d v

= \frac{1}{3} \cdot \int v^{2} d v

Apply the Power Rule

: \frac{v ^{3}}{3}

= \frac{1}{3} \cdot \frac{v ^{3}}{3}

Substitute back

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

Simplify

\frac{1}{3} \cdot \frac{tan ^{3} ( 3 x - 1 )}{3} : \frac{1}{9} tan^{3} (3 x - 1)

= \frac{1}{9} tan^{3} (3 x - 1)

Add a constant to the solution

= \frac{1}{9} tan^{3} (3 x - 1) + C

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What is the integral of tan^2(3x-1)sec^2(3x-1) ?
The integral of tan^2(3x-1)sec^2(3x-1) is 1/9 tan^3(3x-1)+C