What is the integral of 9sec^3(x) ?

The integral of 9sec^3(x) is 9(1/2 sec(x)tan(x)+1/2 ln|tan(x)+sec(x)|)+C

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integral of 9sec^3(x)

\int 9 sec^{3} (x) d x

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

Solution steps

\int 9 sec^{3} (x) d x

Take the constant out:

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

= 9 \cdot \int sec^{3} (x) d x

Apply Integral Reduction

= 9 (\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) ∣

= 9 (\frac{sec ^{2} ( x ) sin ( x )}{2} + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣)

Simplify

9 (\frac{sec ^{2} ( x ) sin ( x )}{2} + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) : 9 (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣)

= 9 (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣)

Add a constant to the solution

= 9 (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + C

\frac{d ^{2}}{d x ^{2}} (1 - x^{2})

\int 23 e^{x} d x

\int \frac{sin ( 2 x )}{13 + cos ^{2} ( x )} d x

y = 4 x, x = 4, y = \frac{4}{x}, x = 0

\frac{\partial}{\partial x} (ln (3 y e^{x y}))

What is the integral of 9sec^3(x) ?
The integral of 9sec^3(x) is 9(1/2 sec(x)tan(x)+1/2 ln|tan(x)+sec(x)|)+C