What is the integral of tan^4(6x)sec^2(6x) ?

The integral of tan^4(6x)sec^2(6x) is 1/30 tan^5(6x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of tan^4(6x)sec^2(6x)

\int tan^{4} (6 x) sec^{2} (6 x) d x

\frac{1}{30} tan^{5} (6 x) + C

Solution steps

\int tan^{4} (6 x) sec^{2} (6 x) d x

Apply u-substitution

: \int tan^{4} (u) sec^{2} (u) \frac{1}{6} d u

= \int tan^{4} (u) sec^{2} (u) \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int tan^{4} (u) sec^{2} (u) d u

Rewrite using trig identities

= \frac{1}{6} \cdot \int (1 + tan^{2} (u)) tan^{4} (u) d u

Apply u-substitution

: \int v^{4} d v

= \frac{1}{6} \cdot \int v^{4} d v

Apply the Power Rule

: \frac{v ^{5}}{5}

= \frac{1}{6} \cdot \frac{v ^{5}}{5}

Substitute back

= \frac{1}{6} \cdot \frac{tan ^{5} ( 6 x )}{5}

Simplify

\frac{1}{6} \cdot \frac{tan ^{5} ( 6 x )}{5} : \frac{1}{30} tan^{5} (6 x)

= \frac{1}{30} tan^{5} (6 x)

Add a constant to the solution

= \frac{1}{30} tan^{5} (6 x) + C

\frac{\partial}{\partial y} (y^{2} - 9)

\frac{d y}{d x} = - \frac{4 x + 4 y}{4 x + 9 y}

x \to 2 + lim (\frac{x - 3}{x ^{2} - 4})

\frac{\partial}{\partial y} (x^{4} + 8 x y^{7})

\int \frac{ln ∣ x ∣}{x} d x

What is the integral of tan^4(6x)sec^2(6x) ?
The integral of tan^4(6x)sec^2(6x) is 1/30 tan^5(6x)+C