What is the integral of ((9-4x^2)^{3/2})/(x^6) ?

The integral of ((9-4x^2)^{3/2})/(x^6) is -((9-4x^2)^{5/2})/(45x^5)+C

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integral of ((9-4x^2)^{3/2})/(x^6)

\int \frac{( 9 - 4 x ^{2} ) ^{\frac{3}{2}}}{x ^{6}} d x

- \frac{( 9 - 4 x ^{2} ) ^{\frac{5}{2}}}{45 x ^{5}} + C

Solution steps

\int \frac{( 9 - 4 x ^{2} ) ^{\frac{3}{2}}}{x ^{6}} d x

Apply Trigonometric Substitution

: \int \frac{32 cos ^{4} ( u )}{9 sin ^{6} ( u )} d u

= \int \frac{32 cos ^{4} ( u )}{9 sin ^{6} ( u )} d u

Take the constant out:

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

= \frac{32}{9} \cdot \int \frac{cos ^{4} ( u )}{sin ^{6} ( u )} d u

Rewrite using trig identities

= \frac{32}{9} \cdot \int cot^{4} (u) csc^{2} (u) d u

Apply u-substitution

: \int - v^{4} d v

= \frac{32}{9} \cdot \int - v^{4} d v

Take the constant out:

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

= \frac{32}{9} (- \int v^{4} d v)

Apply the Power Rule

: \frac{v ^{5}}{5}

= \frac{32}{9} (- \frac{v ^{5}}{5})

Substitute back

= \frac{32}{9} (- \frac{cot ^{5} ( arcsin ( \frac{2}{3} x ) )}{5})

Simplify

\frac{32}{9} (- \frac{cot ^{5} ( arcsin ( \frac{2}{3} x ) )}{5}) : - \frac{( 9 - 4 x ^{2} ) ^{\frac{5}{2}}}{45 x ^{5}}

= - \frac{( 9 - 4 x ^{2} ) ^{\frac{5}{2}}}{45 x ^{5}}

Add a constant to the solution

= - \frac{( 9 - 4 x ^{2} ) ^{\frac{5}{2}}}{45 x ^{5}} + C

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What is the integral of ((9-4x^2)^{3/2})/(x^6) ?
The integral of ((9-4x^2)^{3/2})/(x^6) is -((9-4x^2)^{5/2})/(45x^5)+C