What is the integral of 19tan^3(θ) ?

The integral of 19tan^3(θ) is 19(-ln|sec(θ)|+(sec^2(θ))/2)+C

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integral of 19tan^3(θ)

\int 19 tan^{3} (θ) d θ

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

Solution steps

\int 19 tan^{3} (θ) d θ

Take the constant out:

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

= 19 \cdot \int tan^{3} (θ) d θ

Rewrite using trig identities

= 19 \cdot \int (- 1 + sec^{2} (θ)) tan (θ) d θ

Apply u-substitution

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

= 19 \cdot \int \frac{- 1 + u ^{2}}{u} d u

Expand

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

= 19 \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

= 19 (- \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}

= 19 (- ln ∣ u ∣ + \frac{u ^{2}}{2})

Substitute back

u = sec (θ)

= 19 (- ln ∣ sec (θ) ∣ + \frac{sec ^{2} ( θ )}{2})

Add a constant to the solution

= 19 (- ln ∣ sec (θ) ∣ + \frac{sec ^{2} ( θ )}{2}) + C

\frac{\partial}{\partial y} (x^{5} - 6 x + 4 y^{3} - y^{2})

6 x^{4} - 3

\int x^{4} (x^{5} - 4)^{8} d x

f (x) = \frac{3}{3 x + 1}

What is the integral of 19tan^3(θ) ?
The integral of 19tan^3(θ) is 19(-ln|sec(θ)|+(sec^2(θ))/2)+C