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

The integral of cos(2x)sqrt(4-sin(2x)) is -1/3 (4-sin(2x))^{3/2}+C

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integral of cos(2x)sqrt(4-sin(2x))

\int cos (2 x) 4 - sin (2 x) d x

- \frac{1}{3} (4 - sin (2 x))^{\frac{3}{2}} + C

Solution steps

\int cos (2 x) 4 - sin (2 x) d x

Apply u-substitution

: \int cos (u) 4 - sin (u) \frac{1}{2} d u

= \int cos (u) 4 - sin (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 cos (u) 4 - sin (u) d u

Apply u-substitution

: \int - v d v

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

= \frac{1}{2} (- \frac{2}{3} (4 - sin (2 x))^{\frac{3}{2}})

Simplify

\frac{1}{2} (- \frac{2}{3} (4 - sin (2 x))^{\frac{3}{2}}) : - \frac{1}{3} (4 - sin (2 x))^{\frac{3}{2}}

= - \frac{1}{3} (4 - sin (2 x))^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{3} (4 - sin (2 x))^{\frac{3}{2}} + C

\int 3 e^{5 x + e^{5 x}} d x

\frac{d y}{d x} = \frac{3 x ^{2}}{1 - y}

x^{2} e^{- 5 x}

\int_{\frac{1}{3}}^{5} 6 x ln (3 x) d x

x^{2} \frac{d y}{d x} = y - x y, y (- 1) = - 1

What is the integral of cos(2x)sqrt(4-sin(2x)) ?
The integral of cos(2x)sqrt(4-sin(2x)) is -1/3 (4-sin(2x))^{3/2}+C