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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

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 t)

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

Add a constant to the solution

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

\frac{d ^{2}}{d x ^{2}} (cos^{2} (2 π x))

\int x^{3} d x

ln (x^{2} + 1)

\int_{- 2}^{0} \frac{x}{2} + 1 d x

y^{^{'}} = \frac{x e ^{y}}{y}

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