What is the integral of sin((8t)/9) ?

The integral of sin((8t)/9) is -9/8 cos((8t)/9)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin((8t)/9)

\int sin (\frac{8 t}{9}) d t

- \frac{9}{8} cos (\frac{8 t}{9}) + C

Solution steps

\int sin (\frac{8 t}{9}) d t

Apply u-substitution

: \int sin (u) \frac{9}{8} d u

= \int sin (u) \frac{9}{8} d u

Take the constant out:

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

= \frac{9}{8} \cdot \int sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

= \frac{9}{8} (- cos (u))

Substitute back

u = \frac{8 t}{9}

= \frac{9}{8} (- cos (\frac{8 t}{9}))

Simplify

= - \frac{9}{8} cos (\frac{8 t}{9})

Add a constant to the solution

= - \frac{9}{8} cos (\frac{8 t}{9}) + C

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

\int \frac{x + 7}{x ^{2} + x} d x

\int \frac{3}{1 + 64 x ^{2}} d x

\int t (1 + t^{2}) d t

What is the integral of sin((8t)/9) ?
The integral of sin((8t)/9) is -9/8 cos((8t)/9)+C