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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{- 1 + u ^{2}}{u} d u

= \int \frac{- 1 + u ^{2}}{u} d u

Expand

\frac{- 1 + u ^{2}}{u} : - \frac{1}{u} + u

= \int - \frac{1}{u} + u d u

Apply the Sum Rule:

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

= - \int \frac{1}{u} d u + \int u d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int u d u = \frac{u ^{2}}{2}

= - ln ∣ u ∣ + \frac{u ^{2}}{2}

Substitute back

u = sec (x)

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

Add a constant to the solution

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

\int \frac{7}{( x - 3 ) x ^{2} - 6 x + 5} d x

\frac{\partial}{\partial x} (e^{- x} sin (x))

\frac{d}{d x} (\frac{8}{x - 5})

f (x) = - csc^{2} (x)

(x^{2} - y^{2}) d x + (2 x y) d y = 0

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