What is the integral of sqrt(8-4x^2) ?

The integral of sqrt(8-4x^2) is 2arcsin(x/(sqrt(2)))+sin(2arcsin(x/(sqrt(2))))+C

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integral of sqrt(8-4x^2)

\int 8 - 4 x^{2} d x

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

Solution steps

\int 8 - 4 x^{2} d x

Apply Trigonometric Substitution

: \int 4 cos^{2} (u) d u

= \int 4 cos^{2} (u) d u

Take the constant out:

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

= 4 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 4 \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

= 4 \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

= 4 \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)

= 4 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{2}{2 2} x)

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

Simplify

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

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

Add a constant to the solution

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

\int (x^{2} + x)^{8} (2 x + 1) d x

\int x x^{2} + 1 d x

\int \frac{1}{( x + 2 ) ( x ^{2} + 2 x + 4 ) ^{2}} d x

\int 4 y d y

\int \frac{3 t}{t ^{2} + 1} d t

What is the integral of sqrt(8-4x^2) ?
The integral of sqrt(8-4x^2) is 2arcsin(x/(sqrt(2)))+sin(2arcsin(x/(sqrt(2))))+C