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

\int \frac{3 t}{1 - 9 t ^{4}} d t

\int 3 cos (3 x) d x

f (x) = 5 e^{x} + 7 x, at x = 0

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

\frac{\partial}{\partial t} (e^{2 x} - x cos (t))

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