What is the integral of tan^3(2t)sec^4(2t) ?

The integral of tan^3(2t)sec^4(2t) is (sec^6(2t)}{12}-\frac{sec^4(2t))/8+C

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integral of tan^3(2t)sec^4(2t)

\int tan^{3} (2 t) sec^{4} (2 t) d t

\frac{sec ^{6} ( 2 t )}{12} - \frac{sec ^{4} ( 2 t )}{8} + C

Solution steps

\int tan^{3} (2 t) sec^{4} (2 t) d t

Rewrite using trig identities

= \int (- 1 + sec^{2} (2 t)) tan (2 t) sec^{4} (2 t) d t

Apply u-substitution

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

= \int \frac{u ^{3} ( u ^{2} - 1 )}{2} d u

Expand

\frac{u ^{3} ( u ^{2} - 1 )}{2} : \frac{u ^{5}}{2} - \frac{u ^{3}}{2}

= \int \frac{u ^{5}}{2} - \frac{u ^{3}}{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 ^{5}}{2} d u - \int \frac{u ^{3}}{2} d u

\int \frac{u ^{5}}{2} d u = \frac{u ^{6}}{12}

\int \frac{u ^{3}}{2} d u = \frac{u ^{4}}{8}

= \frac{u ^{6}}{12} - \frac{u ^{4}}{8}

Substitute back

u = sec (2 t)

= \frac{sec ^{6} ( 2 t )}{12} - \frac{sec ^{4} ( 2 t )}{8}

Add a constant to the solution

= \frac{sec ^{6} ( 2 t )}{12} - \frac{sec ^{4} ( 2 t )}{8} + C

\int \frac{4}{x ^{2} - 8 x + 12} d x

y^{^{''}} + y^{^{'}} - 6 y = 4 e^{t}, y (0) = 0, y^{^{'}} (0) = 0

x + x^{- \frac{1}{3}}

\frac{\partial}{\partial x} (3 x^{2} y - 3 x + 3 y^{3})

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

What is the integral of tan^3(2t)sec^4(2t) ?
The integral of tan^3(2t)sec^4(2t) is (sec^6(2t)}{12}-\frac{sec^4(2t))/8+C