What is the integral of sec^2(9x)tan^2(9x) ?

The integral of sec^2(9x)tan^2(9x) is 1/27 tan^3(9x)+C

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integral of sec^2(9x)tan^2(9x)

\int sec^{2} (9 x) tan^{2} (9 x) d x

\frac{1}{27} tan^{3} (9 x) + C

Solution steps

\int sec^{2} (9 x) tan^{2} (9 x) d x

Apply u-substitution

: \int sec^{2} (u) tan^{2} (u) \frac{1}{9} d u

= \int sec^{2} (u) tan^{2} (u) \frac{1}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int sec^{2} (u) tan^{2} (u) d u

Rewrite using trig identities

= \frac{1}{9} \cdot \int (1 + tan^{2} (u)) tan^{2} (u) d u

Apply u-substitution

: \int v^{2} d v

= \frac{1}{9} \cdot \int v^{2} d v

Apply the Power Rule

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

= \frac{1}{9} \cdot \frac{v ^{3}}{3}

Substitute back

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

Simplify

\frac{1}{9} \cdot \frac{tan ^{3} ( 9 x )}{3} : \frac{1}{27} tan^{3} (9 x)

= \frac{1}{27} tan^{3} (9 x)

Add a constant to the solution

= \frac{1}{27} tan^{3} (9 x) + C

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What is the integral of sec^2(9x)tan^2(9x) ?
The integral of sec^2(9x)tan^2(9x) is 1/27 tan^3(9x)+C