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

The integral of tan^3(2x)sec(2x) is -1/2 sec(2x)+(sec^3(2x))/6+C

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

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

- \frac{1}{2} sec (2 x) + \frac{sec ^{3} ( 2 x )}{6} + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

\frac{- 1 + u ^{2}}{2} : - \frac{1}{2} + \frac{u ^{2}}{2}

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

\int \frac{1}{2} d u = \frac{1}{2} u

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

= - \frac{1}{2} u + \frac{u ^{3}}{6}

Substitute back

u = sec (2 x)

= - \frac{1}{2} sec (2 x) + \frac{sec ^{3} ( 2 x )}{6}

Add a constant to the solution

= - \frac{1}{2} sec (2 x) + \frac{sec ^{3} ( 2 x )}{6} + C

\frac{d}{d x} ((1 + \frac{1}{x})^{x})

\frac{d y}{d x} + (3 x y^{3}) = 0, y (0) = 1

\frac{d}{d x} (\frac{3}{6 x + 3})

\int_{0}^{5} \frac{1}{x ^{\frac{21}{22}}} d x

\int \frac{sin ( 2 x )}{42 + cos ^{2} ( x )} d x

What is the integral of tan^3(2x)sec(2x) ?
The integral of tan^3(2x)sec(2x) is -1/2 sec(2x)+(sec^3(2x))/6+C