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

The integral of sec^2(2x)tan^4(2x) is 1/10 tan^5(2x)+C

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

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

\frac{1}{10} tan^{5} (2 x) + C

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int v^{4} d v

= \frac{1}{2} \cdot \int v^{4} d v

Apply the Power Rule

: \frac{v ^{5}}{5}

= \frac{1}{2} \cdot \frac{v ^{5}}{5}

Substitute back

= \frac{1}{2} \cdot \frac{tan ^{5} ( 2 x )}{5}

Simplify

\frac{1}{2} \cdot \frac{tan ^{5} ( 2 x )}{5} : \frac{1}{10} tan^{5} (2 x)

= \frac{1}{10} tan^{5} (2 x)

Add a constant to the solution

= \frac{1}{10} tan^{5} (2 x) + C

\int \frac{1}{5 x ^{2} + 1} d x

x \to + 100 lim (7)

\int \frac{4 + x - 3 x}{x} d x

f (x) = - 2 x^{4} + 5 x^{2} - 3, (1, 0)

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x})

What is the integral of sec^2(2x)tan^4(2x) ?
The integral of sec^2(2x)tan^4(2x) is 1/10 tan^5(2x)+C