What is the integral of 3sec^4(3x) ?

The integral of 3sec^4(3x) is tan(3x)+1/3 tan^3(3x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3sec^4(3x)

\int 3 sec^{4} (3 x) d x

tan (3 x) + \frac{1}{3} tan^{3} (3 x) + C

Solution steps

\int 3 sec^{4} (3 x) d x

Take the constant out:

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

= 3 \cdot \int sec^{4} (3 x) d x

Apply u-substitution

: \int sec^{4} (u) \frac{1}{3} d u

= 3 \cdot \int sec^{4} (u) \frac{1}{3} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int sec^{4} (u) d u

Rewrite using trig identities

= 3 \cdot \frac{1}{3} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) d u

Apply u-substitution

: \int 1 + v^{2} d v

= 3 \cdot \frac{1}{3} \cdot \int 1 + v^{2} d v

Apply the Sum Rule:

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

= 3 \cdot \frac{1}{3} (\int 1 d v + \int v^{2} d v)

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= 3 \cdot \frac{1}{3} (v + \frac{v ^{3}}{3})

Substitute back

= 3 \cdot \frac{1}{3} (tan (3 x) + \frac{tan ^{3} ( 3 x )}{3})

Simplify

3 \cdot \frac{1}{3} (tan (3 x) + \frac{tan ^{3} ( 3 x )}{3}) : tan (3 x) + \frac{1}{3} tan^{3} (3 x)

= tan (3 x) + \frac{1}{3} tan^{3} (3 x)

Add a constant to the solution

= tan (3 x) + \frac{1}{3} tan^{3} (3 x) + C

\int \frac{3 z}{x ^{2} + z ^{2}} d x

\int e^{x} sin (y) - 2 y sin (x) d x

\int \frac{e ^{2 x}}{e ^{2 x} + 1} d x

\int \frac{3 tan ( x )}{cos ( x )} d x

\int \frac{y}{3 y ^{2} + 1} d y

What is the integral of 3sec^4(3x) ?
The integral of 3sec^4(3x) is tan(3x)+1/3 tan^3(3x)+C