What is the integral of tan^3(2x)sec^5(2x) ?

The integral of tan^3(2x)sec^5(2x) is (sec^7(2x))/(14)-(sec^5(2x))/(10)+C

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integral of tan^3(2x)sec^5(2x)

\int tan^{3} (2 x) sec^{5} (2 x) d x

\frac{sec ^{7} ( 2 x )}{14} - \frac{sec ^{5} ( 2 x )}{10} + C

Solution steps

\int tan^{3} (2 x) sec^{5} (2 x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (2 x)) tan (2 x) sec^{5} (2 x) d x

Apply u-substitution

= \int \frac{u ^{4} ( u ^{2} - 1 )}{2} d u

Expand

\frac{u ^{4} ( u ^{2} - 1 )}{2} : \frac{u ^{6}}{2} - \frac{u ^{4}}{2}

= \int \frac{u ^{6}}{2} - \frac{u ^{4}}{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 ^{6}}{2} d u - \int \frac{u ^{4}}{2} d u

\int \frac{u ^{6}}{2} d u = \frac{u ^{7}}{14}

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

= \frac{u ^{7}}{14} - \frac{u ^{5}}{10}

Substitute back

u = sec (2 x)

= \frac{sec ^{7} ( 2 x )}{14} - \frac{sec ^{5} ( 2 x )}{10}

Add a constant to the solution

= \frac{sec ^{7} ( 2 x )}{14} - \frac{sec ^{5} ( 2 x )}{10} + C

\frac{d}{d x} (e^{2 x} cos (4 x))

\int 6 x^{4} - 5 x^{3} + 5 x^{2} + C d x

x \to \infty lim (arctan (1 - x))

x \to 0 + lim (arctan (ln (x)))

\frac{d v}{d x} - \frac{1}{x} v = 8

What is the integral of tan^3(2x)sec^5(2x) ?
The integral of tan^3(2x)sec^5(2x) is (sec^7(2x))/(14)-(sec^5(2x))/(10)+C