What is the integral of 1/((sqrt(16-x^2))^3) ?

The integral of 1/((sqrt(16-x^2))^3) is x/(16sqrt(16-x^2))+C

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integral of 1/((sqrt(16-x^2))^3)

\int \frac{1}{( 16 - x ^{2} ) ^{3}} d x

\frac{x}{16 16 - x ^{2}} + C

Solution steps

\int \frac{1}{( 16 - x ^{2} ) ^{3}} d x

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

= \int \frac{1}{( 16 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{16 cos ^{2} ( u )} d u

= \int \frac{1}{16 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}{16} \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}{16} tan (u)

Substitute back

u = arcsin (\frac{1}{4} x)

= \frac{1}{16} tan (arcsin (\frac{1}{4} x))

Simplify

\frac{1}{16} tan (arcsin (\frac{1}{4} x)) : \frac{x}{16 16 - x ^{2}}

= \frac{x}{16 16 - x ^{2}}

Add a constant to the solution

= \frac{x}{16 16 - x ^{2}} + C

\frac{x ^{2} + 4 x + 3}{x}

ln (- 3 x + 2), - 2

\frac{d}{d x} (\frac{e ^{x - 3}}{x - 2})

x \to - 4 lim (m (x) (3 x + 2) + 2 k)

\int_{5}^{20} (x^{2} x) d x

What is the integral of 1/((sqrt(16-x^2))^3) ?
The integral of 1/((sqrt(16-x^2))^3) is x/(16sqrt(16-x^2))+C