What is the integral of 3x^3sqrt(2-x^2) ?

The integral of 3x^3sqrt(2-x^2) is -2(-x^2+2)^{3/2}+3/5 (-x^2+2)^{5/2}+C

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integral of 3x^3sqrt(2-x^2)

\int 3 x^{3} 2 - x^{2} d x

- 2 (- x^{2} + 2)^{\frac{3}{2}} + \frac{3}{5} (- x^{2} + 2)^{\frac{5}{2}} + C

Solution steps

\int 3 x^{3} 2 - x^{2} d x

Take the constant out:

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

= 3 \cdot \int x^{3} 2 - x^{2} d x

Apply Trigonometric Substitution

: \int 42 sin^{3} (u) cos^{2} (u) d u

= 3 \cdot \int 42 sin^{3} (u) cos^{2} (u) d u

Take the constant out:

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

= 3 \cdot 42 \cdot \int sin^{3} (u) cos^{2} (u) d u

Rewrite using trig identities

= 3 \cdot 42 \cdot \int (1 - cos^{2} (u)) sin (u) cos^{2} (u) d u

Apply u-substitution

: \int - v^{2} (1 - v^{2}) d v

= 3 \cdot 42 \cdot \int - v^{2} (1 - v^{2}) d v

Expand

- v^{2} (1 - v^{2}) : - v^{2} + v^{4}

= 3 \cdot 42 \cdot \int - v^{2} + v^{4} d v

Apply the Sum Rule:

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

= 3 \cdot 42 (- \int v^{2} d v + \int v^{4} d v)

\int v^{2} d v = \frac{v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= 3 \cdot 42 (- \frac{v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= 3 \cdot 42 - \frac{cos ^{3} ( arcsin ( \frac{1}{2} x ) )}{3} + \frac{cos ^{5} ( arcsin ( \frac{1}{2} x ) )}{5}

Simplify

3 \cdot 42 - \frac{cos ^{3} ( arcsin ( \frac{1}{2} x ) )}{3} + \frac{cos ^{5} ( arcsin ( \frac{1}{2} x ) )}{5} : - 2 (- x^{2} + 2)^{\frac{3}{2}} + \frac{3}{5} (- x^{2} + 2)^{\frac{5}{2}}

= - 2 (- x^{2} + 2)^{\frac{3}{2}} + \frac{3}{5} (- x^{2} + 2)^{\frac{5}{2}}

Add a constant to the solution

= - 2 (- x^{2} + 2)^{\frac{3}{2}} + \frac{3}{5} (- x^{2} + 2)^{\frac{5}{2}} + C

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What is the integral of 3x^3sqrt(2-x^2) ?
The integral of 3x^3sqrt(2-x^2) is -2(-x^2+2)^{3/2}+3/5 (-x^2+2)^{5/2}+C