What is the integral of sec^2(3x)tan^4(3x) ?

The integral of sec^2(3x)tan^4(3x) is 1/15 tan^5(3x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sec^2(3x)tan^4(3x)

\int sec^{2} (3 x) tan^{4} (3 x) d x

\frac{1}{15} tan^{5} (3 x) + C

Solution steps

\int sec^{2} (3 x) tan^{4} (3 x) d x

Apply u-substitution

: \int sec^{2} (u) tan^{4} (u) \frac{1}{3} d u

= \int sec^{2} (u) tan^{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

= \frac{1}{3} \cdot \int sec^{2} (u) tan^{4} (u) d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int (1 + tan^{2} (u)) tan^{4} (u) d u

Apply u-substitution

: \int v^{4} d v

= \frac{1}{3} \cdot \int v^{4} d v

Apply the Power Rule

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

= \frac{1}{3} \cdot \frac{v ^{5}}{5}

Substitute back

= \frac{1}{3} \cdot \frac{tan ^{5} ( 3 x )}{5}

Simplify

\frac{1}{3} \cdot \frac{tan ^{5} ( 3 x )}{5} : \frac{1}{15} tan^{5} (3 x)

= \frac{1}{15} tan^{5} (3 x)

Add a constant to the solution

= \frac{1}{15} tan^{5} (3 x) + C

\int \frac{1}{( x + 3 ) ^{\frac{3}{2}}} d x

\int sin (7 x + 3 π) d x

\int (sin (x))^{5} (cos (x))^{4} d x

\int 2 x 4 - x d x

\int x csch (x^{2}) coth (x^{2}) d x

What is the integral of sec^2(3x)tan^4(3x) ?
The integral of sec^2(3x)tan^4(3x) is 1/15 tan^5(3x)+C