Simplification of integral $\int \frac{\sqrt{9-w^2}}{w^2} dw$

Question

I am tasked to find the integral $$\int \frac{\sqrt{9-w^2}}{w^2} dw$$ To solve it, I use the substitution of $w = 3\sin u$, $\frac{dw}{du} = 3\cos u$.

Upon further integration, I arrive at an answer of $-\cot(u) - u + C$, where $C$ is the integration constant. Upon substitution of $u$, I get $$-\cot(\sin^{-1}\frac{w}{3}) - \sin^{-1}\frac{w}{3} + C$$ However, the model answer returns me $$\frac{-\sqrt{9-w^2}}{w^2} - \sin^{-1}\frac{w}{3} + C$$

I'm puzzled how $-\cot(\sin^{-1}\frac{w}{3})$ will end up as $\frac{-\sqrt{9-w^2}}{w^2}$. Is there something I'm missing out on? How do I just replace it with the original expression? Thank you very much!

Answer 1

You wish to simplify:

$$-\cot(\arcsin(\frac{w}{3})) - \arcsin(\frac{w}{3}) + C$$

$$-\frac{\cos(\arcsin(\frac{w}{3})}{\sin(\arcsin(\frac{w}{3})} -\arcsin(\frac{w}{3}) +C$$

Note that $\cos(\arcsin(x)) = \sqrt{1 - \sin^{2}(\arcsin(x))} = \sqrt{1 -x^{2}}$ for $x$ in $\arcsin$'s domain:

$$-\frac{\sqrt{1 - \frac{w^{2}}{9}}}{\frac{w}{3}} - \arcsin(\frac{w}{3}) + C$$

$$-\frac{\sqrt{1 - \frac{w^{2}}{9}}}{\frac{w}{3}}\cdot\frac{3}{3} - \arcsin(\frac{w}{3}) + C$$

$$\boxed{-\frac{\sqrt{9 - w^{2}}}{w} - \arcsin(\frac{w}{3}) + C}$$

Answer 2

Let $\theta=\sin^{-1}\frac{w}3$. Then $\sin\theta=\frac{w}3$, which means $\cot\theta=\frac{\sqrt{9-w^2}}w$.

Draw a right triangle where $w$ is the opposite side and $3$ is the hypotenuse, since $\sin\theta=\frac{w}3$. You will find by the Pythagorean theorem that the adjacent side is $\sqrt{9-w^2}$, which gives the above $\cot\theta$.

Answer 3

Note

$$\cot(\sin^{-1}\frac{w}{3})= \cot(\csc^{-1}\frac{3}{w}) =\sqrt{\csc^2(\csc^{-1}\frac 3w)-1}\\ = \sqrt{(\frac 3w)^2-1}= \frac{\sqrt{9 - w^{2}}}{w} $$

Answer 4

$$\cot(\sin^{-1}(w/3))=\cot \theta , \sin \theta =w/3$$ So $$\cot \theta= \frac{\sqrt{9-w^2}}{w}.$$ Eventually both answers are the same.