What is the integral of 1/((sqrt(9-x^2))^3) ?

The integral of 1/((sqrt(9-x^2))^3) is x/(9sqrt(9-x^2))+C

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integral of 1/((sqrt(9-x^2))^3)

\int \frac{1}{( 9 - x ^{2} ) ^{3}} d x

\frac{x}{9 9 - x ^{2}} + C

Solution steps

\int \frac{1}{( 9 - x ^{2} ) ^{3}} d x

(9 - x^{2})^{3} = (9 - x^{2})^{\frac{3}{2}}

= \int \frac{1}{( 9 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

= \int \frac{1}{9 cos ^{2} ( u )} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{cos ^{2} ( u )} d u

Use the common integral:

\int \frac{1}{cos ^{2} ( u )} d u = tan (u)

= \frac{1}{9} tan (u)

Substitute back

u = arcsin (\frac{1}{3} x)

= \frac{1}{9} tan (arcsin (\frac{1}{3} x))

Simplify

\frac{1}{9} tan (arcsin (\frac{1}{3} x)) : \frac{x}{9 9 - x ^{2}}

= \frac{x}{9 9 - x ^{2}}

Add a constant to the solution

= \frac{x}{9 9 - x ^{2}} + C

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What is the integral of 1/((sqrt(9-x^2))^3) ?
The integral of 1/((sqrt(9-x^2))^3) is x/(9sqrt(9-x^2))+C