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

The integral of cos(4x)sqrt(2-sin(4x)) is -1/6 (2-sin(4x))^{3/2}+C

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

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

- \frac{1}{6} (2 - sin (4 x))^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int - \frac{u}{4} d u

= \int - \frac{u}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int u d u

Apply the Power Rule

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

= - \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 2 - sin (4 x)

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

Simplify

- \frac{1}{4} \cdot \frac{2}{3} (2 - sin (4 x))^{\frac{3}{2}} : - \frac{1}{6} (2 - sin (4 x))^{\frac{3}{2}}

= - \frac{1}{6} (2 - sin (4 x))^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{6} (2 - sin (4 x))^{\frac{3}{2}} + C

\int \frac{1}{( 9 - x ^{2} )} d x

\frac{d}{d x} (4 (x + 4)^{3})

f (x) = x^{2} - 3 x - 2, at x = 2

\frac{d}{d x} (- \frac{b}{x ^{2}})

\frac{d y}{d x} = \frac{cos ( 7 x )}{e ^{7 y}}, y (0) = 0

What is the integral of cos(4x)sqrt(2-sin(4x)) ?
The integral of cos(4x)sqrt(2-sin(4x)) is -1/6 (2-sin(4x))^{3/2}+C