What is the integral of (4sec^3(x)) ?

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

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integral of (4sec^3(x))

\int (4 sec^{3} (x)) d x

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

Solution steps

\int 4 sec^{3} (x) d x

Take the constant out:

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

= 4 \cdot \int sec^{3} (x) d x

Apply Integral Reduction

= 4 (\frac{sec ^{2} ( x ) sin ( x )}{2} + \frac{1}{2} \cdot \int sec (x) d x)

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

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

Simplify

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

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

Add a constant to the solution

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

n = - 4 \sum \infty (\frac{- 4}{9})^{n}

\int 5^{s i n (2 x)} cos (2 x) d x

x \to 0 lim (6 x^{3} + 7)

n = 1 \sum \infty \frac{1}{n}

\int 1 + y e^{x y} d x

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