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

The integral of tan^3(2x)sec^3(2x) is (sec^5(2x)}{10}-\frac{sec^3(2x))/6+C

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

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

\frac{sec ^{5} ( 2 x )}{10} - \frac{sec ^{3} ( 2 x )}{6} + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

\frac{u ^{2} ( u ^{2} - 1 )}{2} : \frac{u ^{4}}{2} - \frac{u ^{2}}{2}

= \int \frac{u ^{4}}{2} - \frac{u ^{2}}{2} 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

= \int \frac{u ^{4}}{2} d u - \int \frac{u ^{2}}{2} d u

\int \frac{u ^{4}}{2} d u = \frac{u ^{5}}{10}

\int \frac{u ^{2}}{2} d u = \frac{u ^{3}}{6}

= \frac{u ^{5}}{10} - \frac{u ^{3}}{6}

Substitute back

u = sec (2 x)

= \frac{sec ^{5} ( 2 x )}{10} - \frac{sec ^{3} ( 2 x )}{6}

Add a constant to the solution

= \frac{sec ^{5} ( 2 x )}{10} - \frac{sec ^{3} ( 2 x )}{6} + C

\frac{d}{d x} (x^{3} - 3 x + 4)

\frac{\partial}{\partial y} (x cos (\frac{y}{z}))

\int \frac{1}{cos ( x ) sin ( x )} d x

\frac{y ^{^{'}}}{3} = x^{2} y, y (0) = - 8

x \frac{d x}{d t} + b t = 0

What is the integral of tan^3(2x)sec^3(2x) ?
The integral of tan^3(2x)sec^3(2x) is (sec^5(2x)}{10}-\frac{sec^3(2x))/6+C