What is the integral of-15sec(-3x)tan(-3x)sec^4(-3x) ?

The integral of-15sec(-3x)tan(-3x)sec^4(-3x) is sec^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-15sec(-3x)tan(-3x)sec^4(-3x)

\int - 15 sec (- 3 x) tan (- 3 x) sec^{4} (- 3 x) d x

sec^{5} (3 x) + C

Solution steps

\int - 15 sec (- 3 x) tan (- 3 x) sec^{4} (- 3 x) d x

Take the constant out:

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

= - 15 \cdot \int sec (- 3 x) tan (- 3 x) sec^{4} (- 3 x) d x

Apply u-substitution

: \int - \frac{1}{3} sec^{5} (u) tan (u) d u

= - 15 \cdot \int - \frac{1}{3} sec^{5} (u) tan (u) d u

Take the constant out:

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

= - 15 (- \frac{1}{3} \cdot \int sec^{5} (u) tan (u) d u)

Apply u-substitution

: \int v^{4} d v

= - 15 (- \frac{1}{3} \cdot \int v^{4} d v)

Apply the Power Rule

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

= - 15 (- \frac{1}{3} \cdot \frac{v ^{5}}{5})

Substitute back

= - 15 (- \frac{1}{3} \cdot \frac{sec ^{5} ( - 3 x )}{5})

Simplify

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

= sec^{5} (3 x)

Add a constant to the solution

= sec^{5} (3 x) + C

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

\int \frac{8}{3 x ^{3} + 7 x ^{2} + 4 x} d x

\int \frac{x}{12 - x ^{4}} d x

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

\int sin (3 x) cos (4 x) cos (5 x) d x

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