What is the integral of tan^5(2x)sec^4(2x) ?

The integral of tan^5(2x)sec^4(2x) is 1/2 ((tan^6(2x))/6+(tan^8(2x))/8)+C

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integral of tan^5(2x)sec^4(2x)

\int tan^{5} (2 x) sec^{4} (2 x) d x

\frac{1}{2} (\frac{tan ^{6} ( 2 x )}{6} + \frac{tan ^{8} ( 2 x )}{8}) + C

Solution steps

\int tan^{5} (2 x) sec^{4} (2 x) d x

Apply u-substitution

: \int tan^{5} (u) sec^{4} (u) \frac{1}{2} d u

= \int tan^{5} (u) sec^{4} (u) \frac{1}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int tan^{5} (u) sec^{4} (u) d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) tan^{5} (u) d u

Apply u-substitution

: \int v^{5} (1 + v^{2}) d v

= \frac{1}{2} \cdot \int v^{5} (1 + v^{2}) d v

Expand

v^{5} (1 + v^{2}) : v^{5} + v^{7}

= \frac{1}{2} \cdot \int v^{5} + v^{7} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int v^{5} d v + \int v^{7} d v)

\int v^{5} d v = \frac{v ^{6}}{6}

\int v^{7} d v = \frac{v ^{8}}{8}

= \frac{1}{2} (\frac{v ^{6}}{6} + \frac{v ^{8}}{8})

Substitute back

= \frac{1}{2} (\frac{tan ^{6} ( 2 x )}{6} + \frac{tan ^{8} ( 2 x )}{8})

Add a constant to the solution

= \frac{1}{2} (\frac{tan ^{6} ( 2 x )}{6} + \frac{tan ^{8} ( 2 x )}{8}) + C

\int \frac{1}{t ^{2} - 4 t + 29} d t

\int 4 - t d t

\int_{0}^{2} (x^{2} - 2)^{4} x d x

y = 4 x, y = \frac{5}{4} x, y = 21 - x^{2}

x \to - 1 - lim (\frac{x ^{3}}{3 x + 3})

What is the integral of tan^5(2x)sec^4(2x) ?
The integral of tan^5(2x)sec^4(2x) is 1/2 ((tan^6(2x))/6+(tan^8(2x))/8)+C