What is the integral of (sqrt((16-9x^2)^3))/(x^6) ?

The integral of (sqrt((16-9x^2)^3))/(x^6) is -((16-9x^2)^{5/2})/(80x^5)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sqrt((16-9x^2)^3))/(x^6)

\int \frac{( 16 - 9 x ^{2} ) ^{3}}{x ^{6}} d x

- \frac{( 16 - 9 x ^{2} ) ^{\frac{5}{2}}}{80 x ^{5}} + C

Solution steps

\int \frac{( 16 - 9 x ^{2} ) ^{3}}{x ^{6}} d x

(16 - 9 x^{2})^{3} = (16 - 9 x^{2})^{\frac{3}{2}},

assuming

(16 - 9 x^{2}) \geq 0

= \int \frac{( 16 - 9 x ^{2} ) ^{\frac{3}{2}}}{x ^{6}} d x

Apply Trigonometric Substitution

: \int \frac{243 cos ^{4} ( u )}{16 sin ^{6} ( u )} d u

= \int \frac{243 cos ^{4} ( u )}{16 sin ^{6} ( u )} d u

Take the constant out:

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

= \frac{243}{16} \cdot \int \frac{cos ^{4} ( u )}{sin ^{6} ( u )} d u

Rewrite using trig identities

= \frac{243}{16} \cdot \int cot^{4} (u) csc^{2} (u) d u

Apply u-substitution

: \int - v^{4} d v

= \frac{243}{16} \cdot \int - v^{4} d v

Take the constant out:

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

= \frac{243}{16} (- \int v^{4} d v)

Apply the Power Rule

: \frac{v ^{5}}{5}

= \frac{243}{16} (- \frac{v ^{5}}{5})

Substitute back

= \frac{243}{16} (- \frac{cot ^{5} ( arcsin ( \frac{3}{4} x ) )}{5})

Simplify

\frac{243}{16} (- \frac{cot ^{5} ( arcsin ( \frac{3}{4} x ) )}{5}) : - \frac{( 16 - 9 x ^{2} ) ^{\frac{5}{2}}}{80 x ^{5}}

= - \frac{( 16 - 9 x ^{2} ) ^{\frac{5}{2}}}{80 x ^{5}}

Add a constant to the solution

= - \frac{( 16 - 9 x ^{2} ) ^{\frac{5}{2}}}{80 x ^{5}} + C

\frac{\partial}{\partial x} (x^{2} cos (x y))

\int x (3 sec^{2} (x) - tan^{2} (x)) d x

\int \frac{x}{x ^{2} + 6 x + 18} d x

\frac{s + 1}{s ( s ^{2} + 4 s + 4 )}

\frac{d}{d x} (x csc (x))

What is the integral of (sqrt((16-9x^2)^3))/(x^6) ?
The integral of (sqrt((16-9x^2)^3))/(x^6) is -((16-9x^2)^{5/2})/(80x^5)+C