What is the integral of 9sec(x)tan(x) ?

The integral of 9sec(x)tan(x) is 9sec(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 9sec(x)tan(x)

\int 9 sec (x) tan (x) d x

9 sec (x) + C

Solution steps

\int 9 sec (x) tan (x) d x

Take the constant out:

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

= 9 \cdot \int sec (x) tan (x) d x

Apply u-substitution

: \int 1 d u

= 9 \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 9 \cdot 1 \cdot u

Substitute back

u = sec (x)

= 9 \cdot 1 \cdot sec (x)

Simplify

= 9 sec (x)

Add a constant to the solution

= 9 sec (x) + C

\int \frac{5 x + 3}{x ^{2} - 4} d x

\int \frac{6 x}{x ^{2} + 4} d x

\frac{x}{1 + x ^{2}}

\int 4 csc^{4} (x) cot (x) d x

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

What is the integral of 9sec(x)tan(x) ?
The integral of 9sec(x)tan(x) is 9sec(x)+C