What is the integral of 1/((16-x^2)^{5/2)} ?

The integral of 1/((16-x^2)^{5/2)} is (x(-x^2+24))/(384(-x^2+16)^{3/2)}+C

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integral of 1/((16-x^2)^{5/2)}

\int \frac{1}{( 16 - x ^{2} ) ^{\frac{5}{2}}} d x

\frac{x ( - x ^{2} + 24 )}{384 ( - x ^{2} + 16 ) ^{\frac{3}{2}}} + C

Solution steps

\int \frac{1}{( 16 - x ^{2} ) ^{\frac{5}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{256 cos ^{4} ( u )} d u

= \int \frac{1}{256 cos ^{4} ( u )} d u

Take the constant out:

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

= \frac{1}{256} \cdot \int \frac{1}{cos ^{4} ( u )} d u

Rewrite using trig identities

= \frac{1}{256} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) d u

Apply u-substitution

: \int 1 + v^{2} d v

= \frac{1}{256} \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

= \frac{1}{256} (\int 1 d v + \int v^{2} d v)

\int 1 d v = v

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

= \frac{1}{256} (v + \frac{v ^{3}}{3})

Substitute back

= \frac{1}{256} (tan (arcsin (\frac{1}{4} x)) + \frac{tan ^{3} ( arcsin ( \frac{1}{4} x ) )}{3})

Simplify

\frac{1}{256} (tan (arcsin (\frac{1}{4} x)) + \frac{tan ^{3} ( arcsin ( \frac{1}{4} x ) )}{3}) : \frac{x ( - x ^{2} + 24 )}{384 ( - x ^{2} + 16 ) ^{\frac{3}{2}}}

= \frac{x ( - x ^{2} + 24 )}{384 ( - x ^{2} + 16 ) ^{\frac{3}{2}}}

Add a constant to the solution

= \frac{x ( - x ^{2} + 24 )}{384 ( - x ^{2} + 16 ) ^{\frac{3}{2}}} + C

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What is the integral of 1/((16-x^2)^{5/2)} ?
The integral of 1/((16-x^2)^{5/2)} is (x(-x^2+24))/(384(-x^2+16)^{3/2)}+C