What is the integral of 4cos(sqrt(2x)) ?

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

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integral of 4cos(sqrt(2x))

\int 4 cos (2 x) d x

4 (2 x sin (2 x) + cos (2 x)) + C

Solution steps

\int 4 cos (2 x) d x

Take the constant out:

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

= 4 \cdot \int cos (2 x) d x

Apply u-substitution

: \int cos (u) \frac{1}{2} d u

= 4 \cdot \int cos (u) \frac{1}{2} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int cos (u) d u

Apply u-substitution

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

= 4 \cdot \frac{1}{2} \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

= 4 \cdot \frac{1}{2} \cdot 2 \cdot \int cos (v) v d v

Apply Integration By Parts

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

= 4 \cdot \frac{1}{2} \cdot 2 (v sin (v) - \int sin (v) d v)

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

= 4 \cdot \frac{1}{2} \cdot 2 (v sin (v) - (- cos (v)))

Substitute back

= 4 \cdot \frac{1}{2} \cdot 2 (2 x sin (2 x) - (- cos (2 x)))

Simplify

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

= 4 (2 x sin (2 x) + cos (2 x))

Add a constant to the solution

= 4 (2 x sin (2 x) + cos (2 x)) + C

\frac{\partial}{\partial y} (x y^{2} + 1)

y^{^{''}} - 43 y^{^{'}} + 12 y = 0

\frac{d}{d x} (\frac{1}{3} ln (e^{x} + 1))

\frac{d}{d x} (- \frac{2 x}{1 - x ^{2}})

f (x) = - x^{2} sin (x) + 4 x cos (x) + 2 sin (x)

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