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

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

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integral of sqrt(2+2cos(4x))

\int 2 + 2 cos (4 x) d x

sin (2 x) + C

Solution steps

\int 2 + 2 cos (4 x) d x

2 + 2 cos (4 x) = 2 1 + cos (4 x)

= \int 2 1 + cos (4 x) d x

Take the constant out:

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

= 2 \cdot \int 1 + cos (4 x) d x

Apply Integral Substitution

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

Take the constant out:

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

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

Rewrite using trig identities

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

Take the constant out:

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

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

cos^{2} (u) = (cos (u)),

assuming

cos (u) \geq 0

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

Use the common integral:

\int cos (u) d u = sin (u)

= 2 \frac{1}{2} 2 sin (u)

Substitute back

u = 2 x

= 2 \frac{1}{2} 2 sin (2 x)

Simplify

2 \frac{1}{2} 2 sin (2 x) : sin (2 x)

= sin (2 x)

Add a constant to the solution

= sin (2 x) + C

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

\frac{d}{d x} (sin (\frac{2 x}{x ^{4} - 4 x}))

n = 0 \sum \infty \frac{9}{n ( n + 3 )}

\int (x^{2}) d x

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

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