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

The integral of 6cos(sqrt(8x)) is 3sqrt(2)sqrt(x)sin(2sqrt(2)sqrt(x))+3/2 cos(2sqrt(2)sqrt(x))+C

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integral of 6cos(sqrt(8x))

\int 6 cos (8 x) d x

32 x sin (22 x) + \frac{3}{2} cos (22 x) + C

Solution steps

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

Apply u-substitution

: \int cos (u) \frac{1}{8} d u

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

Take the constant out:

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

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

Apply u-substitution

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

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

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

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

Substitute back

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

Simplify

6 \cdot \frac{1}{8} \cdot 2 (8 x sin (8 x) - (- cos (8 x))) : 32 x sin (22 x) + \frac{3}{2} cos (22 x)

= 32 x sin (22 x) + \frac{3}{2} cos (22 x)

Add a constant to the solution

= 32 x sin (22 x) + \frac{3}{2} cos (22 x) + C

\frac{1}{z}

\int (3 x^{2} - 2 x + 5) d x

f (x) = - \frac{3}{x}

y^{^{''}} - 5 y^{^{'}} + 6 y = cos (x) e^{- x}

\int \frac{sin ( 2 x )}{32 + cos ^{2} ( x )} d x

What is the integral of 6cos(sqrt(8x)) ?
The integral of 6cos(sqrt(8x)) is 3sqrt(2)sqrt(x)sin(2sqrt(2)sqrt(x))+3/2 cos(2sqrt(2)sqrt(x))+C