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

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

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integral of sec^3(x)-sec(x)

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

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

Solution steps

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

Apply the Sum Rule:

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

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

\int sec^{3} (x) d x = \frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣

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

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

Add similar elements:

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

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

Add a constant to the solution

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

\int_{0}^{1} (\frac{9 t}{1 + t ^{2}}) d t

y^{^{'}} = x (3 - y)

\int (x^{3} (x^{2} - 2 x)^{3}) d x

y = \frac{1}{2} (e^{x} - e^{- x})

925 + 20 x - 0.079 x^{2}

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