What is the integral of 10sqrt(1+cos(4x)) ?

The integral of 10sqrt(1+cos(4x)) is 5sqrt(2)sin(2x)+C

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integral of 10sqrt(1+cos(4x))

\int 10 1 + cos (4 x) d x

52 sin (2 x) + C

Solution steps

\int 10 1 + cos (4 x) d x

Take the constant out:

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

= 10 \cdot \int 1 + cos (4 x) d x

Apply Integral Substitution

= 10 \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

= 10 \cdot \frac{1}{2} \cdot \int 1 + cos (2 u) d u

Rewrite using trig identities

= 10 \cdot \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

= 10 \cdot \frac{1}{2} 2 \cdot \int cos^{2} (u) d u

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

assuming

cos (u) \geq 0

= 10 \cdot \frac{1}{2} 2 \cdot \int cos (u) d u

Use the common integral:

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

= 10 \cdot \frac{1}{2} 2 sin (u)

Substitute back

u = 2 x

= 10 \cdot \frac{1}{2} 2 sin (2 x)

Simplify

10 \cdot \frac{1}{2} 2 sin (2 x) : 52 sin (2 x)

= 52 sin (2 x)

Add a constant to the solution

= 52 sin (2 x) + C

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

\frac{100 p}{0.1 p + 1}

\int 6 x y^{3} d x

\frac{d y}{d x}, 3 x^{3} + 9 x y^{2} = 5 x^{3}

\frac{d}{d x} (\frac{x ^{2} + 3 x + 2}{x + 1})

What is the integral of 10sqrt(1+cos(4x)) ?
The integral of 10sqrt(1+cos(4x)) is 5sqrt(2)sin(2x)+C