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

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

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integral of 2sqrt(1-y^2)

\int 2 1 - y^{2} d y

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

Solution steps

\int 2 1 - y^{2} d y

Take the constant out:

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

= 2 \cdot \int 1 - y^{2} d y

Apply Trigonometric Substitution

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

Rewrite using trig identities

= 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

= 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

= 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)

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

Substitute back

u = arcsin (y)

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

Simplify

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

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

Add a constant to the solution

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

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