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

The integral of sec^2(4x)tan^5(4x) is 1/24 tan^6(4x)+C

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

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

\frac{1}{24} tan^{6} (4 x) + C

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int v^{5} d v

= \frac{1}{4} \cdot \int v^{5} d v

Apply the Power Rule

: \frac{v ^{6}}{6}

= \frac{1}{4} \cdot \frac{v ^{6}}{6}

Substitute back

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

Simplify

\frac{1}{4} \cdot \frac{tan ^{6} ( 4 x )}{6} : \frac{1}{24} tan^{6} (4 x)

= \frac{1}{24} tan^{6} (4 x)

Add a constant to the solution

= \frac{1}{24} tan^{6} (4 x) + C

\int \frac{ln ( x )}{x ^{9}} d x

f (x) = 6 x^{2} - x^{3}, (1, 5)

\frac{\partial}{\partial θ} (sin (θ) cos (φ))

\int 4 x - 6 x y^{2} - 2 d x

t^{2} y^{^{''}} + 19 t y^{^{'}} + 130 y = 0

What is the integral of sec^2(4x)tan^5(4x) ?
The integral of sec^2(4x)tan^5(4x) is 1/24 tan^6(4x)+C