What is the integral of 8sec^3(2x) ?

The integral of 8sec^3(2x) is 4(1/2 sec(2x)tan(2x)+1/2 ln|tan(2x)+sec(2x)|)+C

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integral of 8sec^3(2x)

\int 8 sec^{3} (2 x) d x

4 (\frac{1}{2} sec (2 x) tan (2 x) + \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣) + C

Solution steps

\int 8 sec^{3} (2 x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int sec^{3} (2 x) d x

Apply u-substitution

: \int sec^{3} (u) \frac{1}{2} d u

= 8 \cdot \int sec^{3} (u) \frac{1}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \frac{1}{2} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 8 \cdot \frac{1}{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 8 \cdot \frac{1}{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = 2 x

= 8 \cdot \frac{1}{2} (\frac{sec ^{2} ( 2 x ) sin ( 2 x )}{2} + \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣)

Simplify

8 \cdot \frac{1}{2} (\frac{sec ^{2} ( 2 x ) sin ( 2 x )}{2} + \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣) : 4 (\frac{1}{2} sec (2 x) tan (2 x) + \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣)

= 4 (\frac{1}{2} sec (2 x) tan (2 x) + \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣)

Add a constant to the solution

= 4 (\frac{1}{2} sec (2 x) tan (2 x) + \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣) + C

\int 4 x e^{\frac{2 x}{3}} d x

\int_{1}^{\infty} e^{- 3 x} d x

\int_{3}^{\infty} \frac{1}{x - 2} d x

\int x^{2} + 2 x + 3 d x

\frac{d}{d x} ((x^{2} + y^{2})^{6})

What is the integral of 8sec^3(2x) ?
The integral of 8sec^3(2x) is 4(1/2 sec(2x)tan(2x)+1/2 ln|tan(2x)+sec(2x)|)+C