What is the integral of 5sqrt(1-x^2) ?

The integral of 5sqrt(1-x^2) is 5/2 (arcsin(x)+1/2 sin(2arcsin(x)))+C

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integral of 5sqrt(1-x^2)

\int 5 1 - x^{2} d x

\frac{5}{2} (arcsin (x) + \frac{1}{2} sin (2 arcsin (x))) + C

Solution steps

\int 5 1 - x^{2} d x

Take the constant out:

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

= 5 \cdot \int 1 - x^{2} d x

Apply Trigonometric Substitution

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

= 5 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 5 \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

= 5 \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

= 5 \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)

= 5 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (x)

= 5 \cdot \frac{1}{2} (arcsin (x) + \frac{1}{2} sin (2 arcsin (x)))

Simplify

5 \cdot \frac{1}{2} (arcsin (x) + \frac{1}{2} sin (2 arcsin (x))) : \frac{5}{2} (arcsin (x) + \frac{1}{2} sin (2 arcsin (x)))

= \frac{5}{2} (arcsin (x) + \frac{1}{2} sin (2 arcsin (x)))

Add a constant to the solution

= \frac{5}{2} (arcsin (x) + \frac{1}{2} sin (2 arcsin (x))) + C

\frac{\partial}{\partial x} (3 (x - 1) 2 + 2 (y + 3) 2 + 7)

\int (e^{- 5 x}) d x

x y^{2} \frac{d y}{d x} = x^{3} + y^{3}

\int_{\frac{π}{2}}^{π} 5 cot (\frac{x}{2}) d x

\int - \frac{2 x ^{3} + 24 x}{x ^{4} - 16} d x

What is the integral of 5sqrt(1-x^2) ?
The integral of 5sqrt(1-x^2) is 5/2 (arcsin(x)+1/2 sin(2arcsin(x)))+C