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

The integral of sec^2(2x)tan^2(2x) is 1/6 tan^3(2x)+C

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

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

\frac{1}{6} tan^{3} (2 x) + C

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int v^{2} d v

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

Apply the Power Rule

: \frac{v ^{3}}{3}

= \frac{1}{2} \cdot \frac{v ^{3}}{3}

Substitute back

= \frac{1}{2} \cdot \frac{tan ^{3} ( 2 x )}{3}

Simplify

\frac{1}{2} \cdot \frac{tan ^{3} ( 2 x )}{3} : \frac{1}{6} tan^{3} (2 x)

= \frac{1}{6} tan^{3} (2 x)

Add a constant to the solution

= \frac{1}{6} tan^{3} (2 x) + C

n = 0 \sum \infty \frac{9 ^{n}}{n !}

\frac{d}{d x} (3 x^{2} - 5)

y^{^{'}} = \frac{5 y}{x}

f (x) = 2 x^{2} + x, at x = 0

(1 + x^{2}) (e^{2 t} d t - e^{x} d x) - (1 + x) = 0

What is the integral of sec^2(2x)tan^2(2x) ?
The integral of sec^2(2x)tan^2(2x) is 1/6 tan^3(2x)+C