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

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

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

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

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

Solution steps

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

= \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

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 3

y^{^{''}} - y^{^{'}} - 2 y = e^{2 t}

5 x^{5}

\frac{d}{d t} (\frac{1}{t})

(6 x y - 2 y^{2}) d x + (3 x^{2} - 4 x y) d y = 0

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