What is the integral of 8cos(sqrt(6x)) ?

The integral of 8cos(sqrt(6x)) is 8/3 (sqrt(6)sqrt(x)sin(sqrt(6)sqrt(x))+cos(sqrt(6)sqrt(x)))+C

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integral of 8cos(sqrt(6x))

\int 8 cos (6 x) d x

\frac{8}{3} (6 x sin (6 x) + cos (6 x)) + C

Solution steps

\int 8 cos (6 x) d x

Take the constant out:

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

= 8 \cdot \int cos (6 x) d x

Apply u-substitution

: \int cos (u) \frac{1}{6} d u

= 8 \cdot \int cos (u) \frac{1}{6} d u

Take the constant out:

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

= 8 \cdot \frac{1}{6} \cdot \int cos (u) d u

Apply u-substitution

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

= 8 \cdot \frac{1}{6} \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

= 8 \cdot \frac{1}{6} \cdot 2 \cdot \int cos (v) v d v

Apply Integration By Parts

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

= 8 \cdot \frac{1}{6} \cdot 2 (v sin (v) - \int sin (v) d v)

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

= 8 \cdot \frac{1}{6} \cdot 2 (v sin (v) - (- cos (v)))

Substitute back

= 8 \cdot \frac{1}{6} \cdot 2 (6 x sin (6 x) - (- cos (6 x)))

Simplify

8 \cdot \frac{1}{6} \cdot 2 (6 x sin (6 x) - (- cos (6 x))) : \frac{8}{3} (6 x sin (6 x) + cos (6 x))

= \frac{8}{3} (6 x sin (6 x) + cos (6 x))

Add a constant to the solution

= \frac{8}{3} (6 x sin (6 x) + cos (6 x)) + C

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What is the integral of 8cos(sqrt(6x)) ?
The integral of 8cos(sqrt(6x)) is 8/3 (sqrt(6)sqrt(x)sin(sqrt(6)sqrt(x))+cos(sqrt(6)sqrt(x)))+C