integrate $\int \frac{(16-9x^2)^{\frac{3}{2}}}{x^6}dx$

Question

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

$$\int \frac{(16-9x^2)^{\frac{3}{2}}}{x^6}dx=\int \frac{3\left(\frac{16}{9}-x^2\right)^{\frac{3}{2}}}{x^6}dx$$

$x=\frac{4}{3}\sin\theta$

$dx=\frac{4}{3}\cos\theta d\theta$

$$\int \frac{3\left(\frac{16}{9}-\frac{16}{9}\sin^2\theta\right)^{\frac{3}{2}}}{x^6}d\theta=\int \frac{4(1-\sin^2\theta)^{\frac{3}{2}}}{x^6}d\theta=\int \frac{4(\cos\theta)^3}{x^6}d\theta=4\int \left(\frac{\cos\theta}{\frac{16}{9}\sin\theta}\right)^3d\theta$$

How should I continue?

Answer 1

Notice, you have made mistake while substituting

let $3x=4\sin\theta\implies dx=\frac{4}{3}\cos\theta\ d\theta$ $$\int \frac{(16-9x^2)^{3/2}}{x^6}\ dx$$$$=\int\frac{(16-16\sin^2\theta)^{3/2}}{\left(\frac 43\sin\theta\right)^6} \left(\frac{4}{3}\cos\theta\ d\theta\right)$$ $$=\frac{243}{16}\int \frac{\cos^4\theta}{\sin^6\theta}\ d\theta$$ $$=\frac{243}{16}\int \cot^4\theta\csc^2\theta\ d\theta$$ $$=-\frac{243}{16}\int \cot^4\theta(-\csc^2\theta\ d\theta)$$ $$=-\frac{243}{16}\int \cot^4\theta d(\cot\theta)$$ $$=-\frac{243}{16}\cdot \frac{\cot^5\theta}{5}+C$$ $$=-\frac{243}{80}\cot^5\theta+C$$

Answer 2

Another way using reduction formula:

$$\int(a^2-x^2)^nx^mdx$$

$$=(a^2-x^2)^n\int x^m\ dx-\int\left(\dfrac{d\{(a^2-x^2)^n\}}{dx}\int x^m\ dx\right)dx$$

$$\implies\int(a^2-x^2)^nx^mdx=(a^2-x^2)^n\cdot\dfrac{x^{m+1}}{m+1}+\dfrac{2n}{m+1}\int(a^2-x^2)^{n-1}x^{m+2}\ dx$$

Setting $n=\dfrac32, m=-6$

$$\int(a^2-x^2)^{3/2}x^{-6}dx=-(a^2-x^2)^{3/2}\cdot\dfrac{x^{-5}}5-\dfrac35\int(a^2-x^2)^{1/2}x^{-4}\ dx$$

$n=\dfrac12,m=-4\implies$

$$\int(a^2-x^2)^{1/2}x^{-4}\ dx=-(a^2-x^2)^{1/2}\cdot\dfrac{x^{-3}}3-\dfrac13\int\dfrac1{x^2\sqrt{a^2-x^2}}\ dx$$

Now, $y=\arcsin\dfrac xa,x=a\sin y\implies\cot y=\dfrac{\sqrt{a^2-x^2}}x$

$$\int\dfrac1{x^2\sqrt{a^2-x^2}}\ dx=\int\dfrac{dy}{a^2\csc^2y}dy=c-\dfrac{\cot y}{a^2}$$