What is the integral of sqrt(1+cos(6x)) ?

The integral of sqrt(1+cos(6x)) is (sqrt(2))/3 sin(3x)+C

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integral of sqrt(1+cos(6x))

\int 1 + cos (6 x) d x

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

Solution steps

\int 1 + cos (6 x) d x

Apply Integral Substitution

= \int \frac{1}{2} 1 + cos (3 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 (3 u) d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{2} 1 + cos (\frac{3 v}{2}) 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 (\frac{3 v}{2}) d v

Apply u-substitution

: \int 1 + cos (w) \frac{2}{3} d w

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int 1 + cos (w) \frac{2}{3} d w

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 \frac{2}{3} \cdot \int 1 + cos (w) d w

Apply u-substitution

: \int \frac{2 2}{( 1 + t ^{2} ) 1 + t ^{2}} d t

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot \int \frac{2 2}{( 1 + t ^{2} ) 1 + t ^{2}} d t

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 \frac{2}{3} \cdot 22 \cdot \int \frac{1}{( 1 + t ^{2} ) 1 + t ^{2}} d t

Simplify

(1 + t^{2}) 1 + t^{2} : (1 + t^{2})^{\frac{3}{2}}

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot 22 \cdot \int \frac{1}{( 1 + t ^{2} ) ^{\frac{3}{2}}} d t

Apply Trigonometric Substitution

: \int cos (r) d r

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot 22 \cdot \int cos (r) d r

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot 22 sin (r)

Substitute back

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot 22 sin (arctan (tan (\frac{\frac{3 \cdot 2 \cdot 2 x}{2}}{2})))

Simplify

\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot 22 sin (arctan (tan (\frac{\frac{3 \cdot 2 \cdot 2 x}{2}}{2}))) : \frac{2}{3} sin (3 x)

= \frac{2}{3} sin (3 x)

Add a constant to the solution

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

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What is the integral of sqrt(1+cos(6x)) ?
The integral of sqrt(1+cos(6x)) is (sqrt(2))/3 sin(3x)+C