What is the integral of sin(2x)sqrt(1+2cos(2x)) ?

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

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integral of sin(2x)sqrt(1+2cos(2x))

\int sin (2 x) 1 + 2 cos (2 x) d x

- \frac{1}{6} (1 + 2 cos (2 x))^{\frac{3}{2}} + C

Solution steps

\int sin (2 x) 1 + 2 cos (2 x) d x

Apply u-substitution

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

= \int sin (u) 1 + 2 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

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

Apply u-substitution

: \int - \frac{v}{2} d v

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

Apply the Power Rule

: \frac{2}{3} v^{\frac{3}{2}}

= \frac{1}{2} (- \frac{1}{2} \cdot \frac{2}{3} v^{\frac{3}{2}})

Substitute back

= \frac{1}{2} (- \frac{1}{2} \cdot \frac{2}{3} (1 + 2 cos (2 x))^{\frac{3}{2}})

Simplify

\frac{1}{2} (- \frac{1}{2} \cdot \frac{2}{3} (1 + 2 cos (2 x))^{\frac{3}{2}}) : - \frac{1}{6} (1 + 2 cos (2 x))^{\frac{3}{2}}

= - \frac{1}{6} (1 + 2 cos (2 x))^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{6} (1 + 2 cos (2 x))^{\frac{3}{2}} + C

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