What is the integral of tan^3(6x)sec^3(6x) ?

The integral of tan^3(6x)sec^3(6x) is 1/6 ((sec^5(6x))/5-(sec^3(6x))/3)+C

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integral of tan^3(6x)sec^3(6x)

\int tan^{3} (6 x) sec^{3} (6 x) d x

\frac{1}{6} (\frac{sec ^{5} ( 6 x )}{5} - \frac{sec ^{3} ( 6 x )}{3}) + C

Solution steps

\int tan^{3} (6 x) sec^{3} (6 x) d x

Apply u-substitution

: \int tan^{3} (u) sec^{3} (u) \frac{1}{6} d u

= \int tan^{3} (u) sec^{3} (u) \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int tan^{3} (u) sec^{3} (u) d u

Rewrite using trig identities

= \frac{1}{6} \cdot \int (- 1 + sec^{2} (u)) tan (u) sec^{3} (u) d u

Apply u-substitution

: \int v^{2} (v^{2} - 1) d v

= \frac{1}{6} \cdot \int v^{2} (v^{2} - 1) d v

Expand

v^{2} (v^{2} - 1) : v^{4} - v^{2}

= \frac{1}{6} \cdot \int v^{4} - v^{2} d v

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}{6} (\int v^{4} d v - \int v^{2} d v)

\int v^{4} d v = \frac{v ^{5}}{5}

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

= \frac{1}{6} (\frac{v ^{5}}{5} - \frac{v ^{3}}{3})

Substitute back

= \frac{1}{6} (\frac{sec ^{5} ( 6 x )}{5} - \frac{sec ^{3} ( 6 x )}{3})

Add a constant to the solution

= \frac{1}{6} (\frac{sec ^{5} ( 6 x )}{5} - \frac{sec ^{3} ( 6 x )}{3}) + C

x \to - 2 - lim (\frac{1}{x ^{2} + 2 x})

\frac{d}{d x} (\frac{1}{cosh ( x )})

x \to 3 lim (\frac{10}{x - 3})

h = t^{2} (6 t - 4)^{3}

\frac{\partial}{\partial x} (3 ln (x) - 5)

What is the integral of tan^3(6x)sec^3(6x) ?
The integral of tan^3(6x)sec^3(6x) is 1/6 ((sec^5(6x))/5-(sec^3(6x))/3)+C