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

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

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

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

\frac{1}{4} (\frac{tan ^{6} ( 4 x )}{6} + \frac{tan ^{8} ( 4 x )}{8}) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= \frac{1}{4} \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}{4} \cdot \int v^{5} (1 + v^{2}) d v

Expand

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

= \frac{1}{4} \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}{4} (\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}{4} (\frac{v ^{6}}{6} + \frac{v ^{8}}{8})

Substitute back

= \frac{1}{4} (\frac{tan ^{6} ( 4 x )}{6} + \frac{tan ^{8} ( 4 x )}{8})

Add a constant to the solution

= \frac{1}{4} (\frac{tan ^{6} ( 4 x )}{6} + \frac{tan ^{8} ( 4 x )}{8}) + C

x \to - \infty lim (- x^{2})

\frac{1}{( s ^{2} - 2 s + 5 )}

f (x) = 25 - x, at x = 0

\int (3^{- x}) d x

\frac{d y}{d x}, y = x^{2} + 3

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