What is the integral of 8/9 sec^2(x/9) ?

The integral of 8/9 sec^2(x/9) is 8tan(x/9)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 8/9 sec^2(x/9)

\int \frac{8}{9} sec^{2} (\frac{x}{9}) d x

8 tan (\frac{x}{9}) + C

Solution steps

\int \frac{8}{9} sec^{2} (\frac{x}{9}) d x

Take the constant out:

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

= \frac{8}{9} \cdot \int sec^{2} (\frac{x}{9}) d x

Apply u-substitution

: \int sec^{2} (u) \cdot 9 d u

= \frac{8}{9} \cdot \int sec^{2} (u) \cdot 9 d u

Take the constant out:

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

= \frac{8}{9} \cdot 9 \cdot \int sec^{2} (u) d u

Use the common integral:

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

= \frac{8}{9} \cdot 9 tan (u)

Substitute back

u = \frac{x}{9}

= \frac{8}{9} \cdot 9 tan (\frac{x}{9})

Simplify

\frac{8}{9} \cdot 9 tan (\frac{x}{9}) : 8 tan (\frac{x}{9})

= 8 tan (\frac{x}{9})

Add a constant to the solution

= 8 tan (\frac{x}{9}) + C

\int cos^{11} (θ) sin (θ) d θ

\int \frac{x ^{4} e ^{x^{5}}}{5} d x

\int e^{x} 6 x d x

\int \frac{4 x - 8}{x ^{2} - 4 x + 5} d x

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

What is the integral of 8/9 sec^2(x/9) ?
The integral of 8/9 sec^2(x/9) is 8tan(x/9)+C