What is the integral of sqrt(1-4w^2) ?

The integral of sqrt(1-4w^2) is 1/4 (arcsin(2w)+1/2 sin(2arcsin(2w)))+C

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integral of sqrt(1-4w^2)

\int 1 - 4 w^{2} d w

\frac{1}{4} (arcsin (2 w) + \frac{1}{2} sin (2 arcsin (2 w))) + C

Solution steps

\int 1 - 4 w^{2} d w

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d u = u

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

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

Substitute back

u = arcsin (2 w)

= \frac{1}{2} \cdot \frac{1}{2} (arcsin (2 w) + \frac{1}{2} sin (2 arcsin (2 w)))

Simplify

\frac{1}{2} \cdot \frac{1}{2} (arcsin (2 w) + \frac{1}{2} sin (2 arcsin (2 w))) : \frac{1}{4} (arcsin (2 w) + \frac{1}{2} sin (2 arcsin (2 w)))

= \frac{1}{4} (arcsin (2 w) + \frac{1}{2} sin (2 arcsin (2 w)))

Add a constant to the solution

= \frac{1}{4} (arcsin (2 w) + \frac{1}{2} sin (2 arcsin (2 w))) + C

\int \frac{sin ( 2 x )}{cos ^{3} ( 2 x )} d x

x, \frac{1}{x ^{2}}, 1, 3

\frac{d y}{d x} = - 5 y x, y (0) = 50

x 2^{x}

\int sin^{4} (x) \cdot cos^{5} (x) d x

What is the integral of sqrt(1-4w^2) ?
The integral of sqrt(1-4w^2) is 1/4 (arcsin(2w)+1/2 sin(2arcsin(2w)))+C