What is the integral of 6cos(sqrt(3x)) ?

The integral of 6cos(sqrt(3x)) is 4(sqrt(3)sqrt(x)sin(sqrt(3)sqrt(x))+cos(sqrt(3)sqrt(x)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6cos(sqrt(3x))

\int 6 cos (3 x) d x

4 (3 x sin (3 x) + cos (3 x)) + C

Solution steps

\int 6 cos (3 x) d x

Take the constant out:

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

= 6 \cdot \int cos (3 x) d x

Apply u-substitution

: \int cos (u) \frac{1}{3} d u

= 6 \cdot \int cos (u) \frac{1}{3} d u

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \int cos (u) d u

Apply u-substitution

: \int cos (v) \cdot 2 v d v

= 6 \cdot \frac{1}{3} \cdot \int cos (v) \cdot 2 v d v

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot 2 \cdot \int cos (v) v d v

Apply Integration By Parts

: v sin (v) - \int sin (v) d v

= 6 \cdot \frac{1}{3} \cdot 2 (v sin (v) - \int sin (v) d v)

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

= 6 \cdot \frac{1}{3} \cdot 2 (v sin (v) - (- cos (v)))

Substitute back

= 6 \cdot \frac{1}{3} \cdot 2 (3 x sin (3 x) - (- cos (3 x)))

Simplify

6 \cdot \frac{1}{3} \cdot 2 (3 x sin (3 x) - (- cos (3 x))) : 4 (3 x sin (3 x) + cos (3 x))

= 4 (3 x sin (3 x) + cos (3 x))

Add a constant to the solution

= 4 (3 x sin (3 x) + cos (3 x)) + C

x \to 0 lim (\frac{sin ( 10 x )}{sin ( 3 x )})

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

\frac{\partial}{\partial x} (x^{3} + x^{2} y^{2} + y^{3} + x + y)

\frac{\partial}{\partial x} (ln (x + 2 y + 9 z))

- (1 - 4 \cdot z)

What is the integral of 6cos(sqrt(3x)) ?
The integral of 6cos(sqrt(3x)) is 4(sqrt(3)sqrt(x)sin(sqrt(3)sqrt(x))+cos(sqrt(3)sqrt(x)))+C