What is the integral of sqrt(4-x^2)(16-4x^2) ?

The integral of sqrt(4-x^2)(16-4x^2) is 24arcsin(1/2 x)+16sin(2arcsin(1/2 x))+2sin(4arcsin(1/2 x))+C

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integral of sqrt(4-x^2)(16-4x^2)

\int 4 - x^{2} (16 - 4 x^{2}) d x

24 arcsin (\frac{1}{2} x) + 16 sin (2 arcsin (\frac{1}{2} x)) + 2 sin (4 arcsin (\frac{1}{2} x)) + C

Solution steps

\int 4 - x^{2} (16 - 4 x^{2}) d x

Expand

4 - x^{2} (16 - 4 x^{2}) : 16 4 - x^{2} - 4 x^{2} 4 - x^{2}

= \int 16 4 - x^{2} - 4 x^{2} 4 - x^{2} d x

Apply the Sum Rule:

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

= \int 16 4 - x^{2} d x - \int 4 x^{2} 4 - x^{2} d x

\int 16 4 - x^{2} d x = 32 (arcsin (\frac{1}{2} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} x)))

\int 4 x^{2} 4 - x^{2} d x = 8 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x)))

= 32 (arcsin (\frac{1}{2} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} x))) - 8 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x)))

Simplify

32 (arcsin (\frac{1}{2} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} x))) - 8 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x))) : 24 arcsin (\frac{1}{2} x) + 16 sin (2 arcsin (\frac{1}{2} x)) + 2 sin (4 arcsin (\frac{1}{2} x))

= 24 arcsin (\frac{1}{2} x) + 16 sin (2 arcsin (\frac{1}{2} x)) + 2 sin (4 arcsin (\frac{1}{2} x))

Add a constant to the solution

= 24 arcsin (\frac{1}{2} x) + 16 sin (2 arcsin (\frac{1}{2} x)) + 2 sin (4 arcsin (\frac{1}{2} x)) + C

\int cos (x z) d x

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\int (x - 5)^{2} d x

\int_{1}^{3} x^{3} ln (x) d x

\frac{d}{d x} (\frac{3 x}{2 x + 1})

What is the integral of sqrt(4-x^2)(16-4x^2) ?
The integral of sqrt(4-x^2)(16-4x^2) is 24arcsin(1/2 x)+16sin(2arcsin(1/2 x))+2sin(4arcsin(1/2 x))+C