What is the integral of 5/((9-x^2)^{5/2)} ?

The integral of 5/((9-x^2)^{5/2)} is (5(-2x^3+27x))/(243(-x^2+9)^{3/2)}+C

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integral of 5/((9-x^2)^{5/2)}

\int \frac{5}{( 9 - x ^{2} ) ^{\frac{5}{2}}} d x

\frac{5 ( - 2 x ^{3} + 27 x )}{243 ( - x ^{2} + 9 ) ^{\frac{3}{2}}} + C

Solution steps

\int \frac{5}{( 9 - x ^{2} ) ^{\frac{5}{2}}} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{( 9 - x ^{2} ) ^{\frac{5}{2}}} d x

Apply Trigonometric Substitution

= 5 \cdot \int \frac{1}{81 cos ^{4} ( u )} d u

Take the constant out:

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

= 5 \cdot \frac{1}{81} \cdot \int \frac{1}{cos ^{4} ( u )} d u

Rewrite using trig identities

= 5 \cdot \frac{1}{81} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) d u

Apply u-substitution

= 5 \cdot \frac{1}{81} \cdot \int 1 + 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

= 5 \cdot \frac{1}{81} (\int 1 d v + \int v^{2} d v)

\int 1 d v = v

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

= 5 \cdot \frac{1}{81} (v + \frac{v ^{3}}{3})

Substitute back

= 5 \cdot \frac{1}{81} (tan (arcsin (\frac{1}{3} x)) + \frac{tan ^{3} ( arcsin ( \frac{1}{3} x ) )}{3})

Simplify

5 \cdot \frac{1}{81} (tan (arcsin (\frac{1}{3} x)) + \frac{tan ^{3} ( arcsin ( \frac{1}{3} x ) )}{3}) : \frac{5 ( - 2 x ^{3} + 27 x )}{243 ( - x ^{2} + 9 ) ^{\frac{3}{2}}}

= \frac{5 ( - 2 x ^{3} + 27 x )}{243 ( - x ^{2} + 9 ) ^{\frac{3}{2}}}

Add a constant to the solution

= \frac{5 ( - 2 x ^{3} + 27 x )}{243 ( - x ^{2} + 9 ) ^{\frac{3}{2}}} + C

What is the integral of 5/((9-x^2)^{5/2)} ?
The integral of 5/((9-x^2)^{5/2)} is (5(-2x^3+27x))/(243(-x^2+9)^{3/2)}+C