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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{( e ^{x} ( 2 x + 3 ) )}{x ^{2} + 1}

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

\frac{8}{x}

\frac{d x}{d t} + 5 x = 50

\frac{d}{d x} (\frac{7 e ^{x}}{2 e ^{x} + 1})

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