What is the integral of sqrt(16x^4-24x^2+9) ?

The integral of sqrt(16x^4-24x^2+9) is 4/3 arccos^3(cos(x))+3arcsin(cos(x))+C

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integral of sqrt(16x^4-24x^2+9)

\int 16 x^{4} - 24 x^{2} + 9 d x

\frac{4}{3} arccos^{3} (cos (x)) + 3 arcsin (cos (x)) + C

Solution steps

\int 16 x^{4} - 24 x^{2} + 9 d x

Apply u-substitution

: \int - \frac{( 2 arccos ( u ) + 3 ) ( 2 arccos ( u ) - 3 )}{1 - u ^{2}} d u

= \int - \frac{( 2 arccos ( u ) + 3 ) ( 2 arccos ( u ) - 3 )}{1 - u ^{2}} d u

Take the constant out:

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

= - \int \frac{( 2 arccos ( u ) + 3 ) ( 2 arccos ( u ) - 3 )}{1 - u ^{2}} d u

Expand

\frac{( 2 arccos ( u ) + 3 ) ( 2 arccos ( u ) - 3 )}{1 - u ^{2}} : \frac{4 arccos ^{2} ( u )}{1 - u ^{2}} - \frac{3}{1 - u ^{2}}

= - \int \frac{4 arccos ^{2} ( u )}{1 - u ^{2}} - \frac{3}{1 - u ^{2}} 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

= - (\int \frac{4 arccos ^{2} ( u )}{1 - u ^{2}} d u - \int \frac{3}{1 - u ^{2}} d u)

\int \frac{4 arccos ^{2} ( u )}{1 - u ^{2}} d u = - \frac{4}{3} arccos^{3} (u)

\int \frac{3}{1 - u ^{2}} d u = 3 arcsin (u)

= - (- \frac{4}{3} arccos^{3} (u) - 3 arcsin (u))

Substitute back

u = cos (x)

= - (- \frac{4}{3} arccos^{3} (cos (x)) - 3 arcsin (cos (x)))

Simplify

= \frac{4}{3} arccos^{3} (cos (x)) + 3 arcsin (cos (x))

Add a constant to the solution

= \frac{4}{3} arccos^{3} (cos (x)) + 3 arcsin (cos (x)) + C

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What is the integral of sqrt(16x^4-24x^2+9) ?
The integral of sqrt(16x^4-24x^2+9) is 4/3 arccos^3(cos(x))+3arcsin(cos(x))+C