What is the integral of 2cos(sqrt(9x)) ?

The integral of 2cos(sqrt(9x)) is 4/9 (3sqrt(x)sin(3sqrt(x))+cos(3sqrt(x)))+C

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integral of 2cos(sqrt(9x))

\int 2 cos (9 x) d x

\frac{4}{9} (3 x sin (3 x) + cos (3 x)) + C

Solution steps

\int 2 cos (9 x) d x

Take the constant out:

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

= 2 \cdot \int cos (9 x) d x

Apply u-substitution

: \int cos (u) \frac{1}{9} d u

= 2 \cdot \int cos (u) \frac{1}{9} d u

Take the constant out:

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

= 2 \cdot \frac{1}{9} \cdot \int cos (u) d u

Apply u-substitution

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

= 2 \cdot \frac{1}{9} \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

= 2 \cdot \frac{1}{9} \cdot 2 \cdot \int cos (v) v d v

Apply Integration By Parts

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

= 2 \cdot \frac{1}{9} \cdot 2 (v sin (v) - \int sin (v) d v)

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

= 2 \cdot \frac{1}{9} \cdot 2 (v sin (v) - (- cos (v)))

Substitute back

= 2 \cdot \frac{1}{9} \cdot 2 (9 x sin (9 x) - (- cos (9 x)))

Simplify

2 \cdot \frac{1}{9} \cdot 2 (9 x sin (9 x) - (- cos (9 x))) : \frac{4}{9} (3 x sin (3 x) + cos (3 x))

= \frac{4}{9} (3 x sin (3 x) + cos (3 x))

Add a constant to the solution

= \frac{4}{9} (3 x sin (3 x) + cos (3 x)) + C

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What is the integral of 2cos(sqrt(9x)) ?
The integral of 2cos(sqrt(9x)) is 4/9 (3sqrt(x)sin(3sqrt(x))+cos(3sqrt(x)))+C