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

The integral of (x^5)/(sqrt(1-2x^2)) is -1/30 (3x^4+2x^2+2)sqrt(-2x^2+1)+C

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integral of (x^5)/(sqrt(1-2x^2))

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

- \frac{1}{30} (3 x^{4} + 2 x^{2} + 2) - 2 x^{2} + 1 + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{sin ^{5} ( u )}{8} d u

= \int \frac{sin ^{5} ( u )}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int sin^{5} (u) d u

Rewrite using trig identities

= \frac{1}{8} \cdot \int (1 - cos^{2} (u))^{2} sin (u) d u

Apply u-substitution

: \int - 1 + 2 v^{2} - v^{4} d v

= \frac{1}{8} \cdot \int - 1 + 2 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

= \frac{1}{8} (- \int 1 d v + \int 2 v^{2} d v - \int v^{4} d v)

\int 1 d v = v

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

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

= \frac{1}{8} (- v + \frac{2 v ^{3}}{3} - \frac{v ^{5}}{5})

Substitute back

= \frac{1}{8} (- cos (arcsin (2 x)) + \frac{2 cos ^{3} ( arcsin ( 2 x ) )}{3} - \frac{cos ^{5} ( arcsin ( 2 x ) )}{5})

Simplify

\frac{1}{8} (- cos (arcsin (2 x)) + \frac{2 cos ^{3} ( arcsin ( 2 x ) )}{3} - \frac{cos ^{5} ( arcsin ( 2 x ) )}{5}) : - \frac{1}{30} (3 x^{4} + 2 x^{2} + 2) - 2 x^{2} + 1

= - \frac{1}{30} (3 x^{4} + 2 x^{2} + 2) - 2 x^{2} + 1

Add a constant to the solution

= - \frac{1}{30} (3 x^{4} + 2 x^{2} + 2) - 2 x^{2} + 1 + C

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What is the integral of (x^5)/(sqrt(1-2x^2)) ?
The integral of (x^5)/(sqrt(1-2x^2)) is -1/30 (3x^4+2x^2+2)sqrt(-2x^2+1)+C