What is the integral of (sqrt(1-x^2))/(x^6) ?

The integral of (sqrt(1-x^2))/(x^6) is -((1-x^2)^{5/2})/(5x^5)-((1-x^2)^{3/2})/(3x^3)+C

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integral of (sqrt(1-x^2))/(x^6)

\int \frac{1 - x ^{2}}{x ^{6}} d x

- \frac{( 1 - x ^{2} ) ^{\frac{5}{2}}}{5 x ^{5}} - \frac{( 1 - x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}} + C

Solution steps

\int \frac{1 - x ^{2}}{x ^{6}} d x

Apply Trigonometric Substitution

: \int \frac{cos ^{2} ( u )}{sin ^{6} ( u )} d u

= \int \frac{cos ^{2} ( u )}{sin ^{6} ( u )} d u

Rewrite using trig identities

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Expand

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

= - \int v^{4} + v^{2} 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

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

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

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

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

Substitute back

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

Simplify

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

= - \frac{( 1 - x ^{2} ) ^{\frac{5}{2}}}{5 x ^{5}} - \frac{( 1 - x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}}

Add a constant to the solution

= - \frac{( 1 - x ^{2} ) ^{\frac{5}{2}}}{5 x ^{5}} - \frac{( 1 - x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}} + C

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What is the integral of (sqrt(1-x^2))/(x^6) ?
The integral of (sqrt(1-x^2))/(x^6) is -((1-x^2)^{5/2})/(5x^5)-((1-x^2)^{3/2})/(3x^3)+C