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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= d \cdot \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

= d (- \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}

= d (- ln ∣ u ∣ + \frac{u ^{2}}{2})

Substitute back

u = sec (x)

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

Add a constant to the solution

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

\int 25 - x^{2} d x

z \to 0 lim (\frac{8}{z ^{3}} + \frac{2}{z} + 2)

y = \frac{9}{x ^{2}}, y = 10 - x^{2}, y = 12

\frac{d y}{d x} - \frac{y}{2 x} = 3 x

x (x^{2} - 11 x + 35), x^{2}, 0, 7

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