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

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

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

Apply the Power Rule

= \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

x \to π lim (\frac{x}{4} - \frac{π}{2})

\int \frac{s ^{2}}{s ^{3} + 5} d s

\int_{0}^{1} 2 x^{2} d x

\int \frac{1}{x ^{2} - 2 x - 8} d x

\frac{d}{d x} (ln (\frac{1}{6 x}))

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