What is the integral of sqrt(1+cos(8x)) ?

The integral of sqrt(1+cos(8x)) is 1/(2sqrt(2))sin(4x)+C

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integral of sqrt(1+cos(8x))

\int 1 + cos (8 x) d x

\frac{1}{2 2} sin (4 x) + C

Solution steps

\int 1 + cos (8 x) d x

Apply Integral Substitution

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Rewrite using trig identities

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

Take the constant out:

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

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

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

assuming

cos (v) \geq 0

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{2 2} sin (4 x) + C

\frac{d}{d x} (cos (arctan (5 x)))

n = 0 \sum \infty n^{n} (x - 3)^{n}

x \to 2 lim (\frac{3}{x + 2})

\frac{s}{( s - 2 ) ( s - 3 ) ( s - 6 )}

\frac{d}{d x} (\frac{( x ^{2} + 1 )}{( x + 4 )})

What is the integral of sqrt(1+cos(8x)) ?
The integral of sqrt(1+cos(8x)) is 1/(2sqrt(2))sin(4x)+C