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

The integral of sec^4(2x) is 1/2 (tan(2x)+(tan^3(2x))/3)+C

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integral of sec^4(2x)

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

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

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int 1 + v^{2} d v

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int 1 d v + \int v^{2} d v)

\int 1 d v = v

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

= \frac{1}{2} (v + \frac{v ^{3}}{3})

Substitute back

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

Add a constant to the solution

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

x \to 4 - lim (e^{\frac{1}{x - 4}})

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

\frac{d}{d x} (\frac{x ^{2}}{4} - \frac{ln ( x )}{2})

x \to - 9 lim (- 12)

f (x) = - sin (x), 0, 2 π

What is the integral of sec^4(2x) ?
The integral of sec^4(2x) is 1/2 (tan(2x)+(tan^3(2x))/3)+C