What is the integral of ((1-r^2)^{5/2})/(r^8) ?

The integral of ((1-r^2)^{5/2})/(r^8) is -((1-r^2)^{7/2})/(7r^7)+C

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integral of ((1-r^2)^{5/2})/(r^8)

\int \frac{( 1 - r ^{2} ) ^{\frac{5}{2}}}{r ^{8}} d r

- \frac{( 1 - r ^{2} ) ^{\frac{7}{2}}}{7 r ^{7}} + C

Solution steps

\int \frac{( 1 - r ^{2} ) ^{\frac{5}{2}}}{r ^{8}} d r

Apply Trigonometric Substitution

: \int \frac{cos ^{6} ( u )}{sin ^{8} ( u )} d u

= \int \frac{cos ^{6} ( u )}{sin ^{8} ( u )} d u

Rewrite using trig identities

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

Apply u-substitution

: \int - v^{6} d v

= \int - v^{6} d v

Take the constant out:

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

= - \int v^{6} d v

Apply the Power Rule

: \frac{v ^{7}}{7}

= - \frac{v ^{7}}{7}

Substitute back

= - \frac{cot ^{7} ( arcsin ( r ) )}{7}

Simplify

- \frac{cot ^{7} ( arcsin ( r ) )}{7} : - \frac{( 1 - r ^{2} ) ^{\frac{7}{2}}}{7 r ^{7}}

= - \frac{( 1 - r ^{2} ) ^{\frac{7}{2}}}{7 r ^{7}}

Add a constant to the solution

= - \frac{( 1 - r ^{2} ) ^{\frac{7}{2}}}{7 r ^{7}} + C

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What is the integral of ((1-r^2)^{5/2})/(r^8) ?
The integral of ((1-r^2)^{5/2})/(r^8) is -((1-r^2)^{7/2})/(7r^7)+C