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

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

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integral of sec(x)tan^2(x)

integral

sec (x) tan^{2} (x)

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

Solution steps

\int sec (x) tan^{2} (x) d x

Rewrite using trig identities

= \int sec (x) (- 1 + sec^{2} (x)) d x

Expand

sec (x) (- 1 + sec^{2} (x)) : - sec (x) + sec^{3} (x)

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

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

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

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

Simplify

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

ln (x + 1)

2 + 3 t

\frac{( e ^{- x} ) ^{2}}{2}

e^{- x} sin (2 x)

\frac{- 2}{x}

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