What is the integral of 11tan^3(x) ?

The integral of 11tan^3(x) is 11(-ln|sec(x)|+(sec^2(x))/2)+C

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integral of 11tan^3(x)

\int 11 tan^{3} (x) d x

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

Solution steps

\int 11 tan^{3} (x) d x

Take the constant out:

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

= 11 \cdot \int tan^{3} (x) d x

Rewrite using trig identities

= 11 \cdot \int (- 1 + sec^{2} (x)) tan (x) d x

Apply u-substitution

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

= 11 \cdot \int \frac{- 1 + u ^{2}}{u} d u

Expand

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

= 11 \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

= 11 (- \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}

= 11 (- ln ∣ u ∣ + \frac{u ^{2}}{2})

Substitute back

u = sec (x)

= 11 (- ln ∣ sec (x) ∣ + \frac{sec ^{2} ( x )}{2})

Add a constant to the solution

= 11 (- ln ∣ sec (x) ∣ + \frac{sec ^{2} ( x )}{2}) + C

\frac{d}{d x} (ln (x^{4} - 4 x))

\int_{0}^{2} π (x + 1)^{2} d x

\frac{d}{d x} (x^{4} + 3 x)

\int \frac{lo g _{3} ( x )}{x} d x

y^{^{'}} + (\frac{1 + x}{x}) y = 0

What is the integral of 11tan^3(x) ?
The integral of 11tan^3(x) is 11(-ln|sec(x)|+(sec^2(x))/2)+C