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

The integral of sec^4(2x)tan(2x) is 1/8 sec^4(2x)+C

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

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

\frac{1}{8} sec^{4} (2 x) + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int v^{3} d v

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

Apply the Power Rule

: \frac{v ^{4}}{4}

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

Substitute back

= \frac{1}{2} \cdot \frac{sec ^{4} ( 2 x )}{4}

Simplify

\frac{1}{2} \cdot \frac{sec ^{4} ( 2 x )}{4} : \frac{1}{8} sec^{4} (2 x)

= \frac{1}{8} sec^{4} (2 x)

Add a constant to the solution

= \frac{1}{8} sec^{4} (2 x) + C

e^{- \frac{x}{2}}

\int 6 y^{- 4} d y

5 x - x^{2}, (4, 4)

\int 5 - 2 x d x

t \to 0 lim (cos (t))

What is the integral of sec^4(2x)tan(2x) ?
The integral of sec^4(2x)tan(2x) is 1/8 sec^4(2x)+C