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I am trying to solve $$\int\frac{\sqrt{9-x^2}}{x^2}\mathrm dx$$

My answer is slightly different to the memo:

$x=3\sin\theta\quad\iff\quad\theta=\arcsin\left(\frac x 3\right)\\ \text dx=3\cos\theta\ \text d\theta\\$ $\begin{align}I&=\int\frac{3\sqrt{1-\sin^2\theta}}{3\sin^2\theta}\cdot3\cos\theta\ \text d\theta=3\int\frac{\cos^2\theta}{\sin^2\theta}\ \text d\theta=3\int\cot^2\theta\ \text d \theta\\ &=3\int\csc^2\theta\ \text d\theta - 3\int\text d \theta\\ &=-3\cot\theta-3\theta+C\\ &=-\frac{\sqrt{1-\left(\frac x 3\right)^2}}{\frac x 3}-3\arcsin\left(\frac x 3\right)+C\\ &=-\frac{3\sqrt{9-x^2}}{3x}-3\arcsin\left(\frac x 3\right)+C\\ &=-\frac{\sqrt{9-x^2}}x-3\arcsin\left(\frac x 3\right)+C \end{align}$

and the memo has $$-\frac{\sqrt{9-x^2}}x-\arcsin\left(\frac x 3\right)+C$$

Your help is appreciated!

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    $\begingroup$ Third line from top - denominator should be $9\sin^2\theta$, NOT $3\sin^2\theta$ $\endgroup$ – AgentS Aug 19 '14 at 13:27
  • $\begingroup$ When you replace $x=3\sin{\theta}$ in the denominator should be $9\sin^2{\theta}$ $\endgroup$ – rlartiga Aug 19 '14 at 13:29
  • $\begingroup$ Yeah, what they said (you lost a 3 downstairs) $\endgroup$ – amcalde Aug 19 '14 at 13:36
  • $\begingroup$ Thanks. In that case, I retrospectively made a mistake with the first term as well ($-3\cot\theta$) $\endgroup$ – ahorn Aug 19 '14 at 13:40
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$$\int\frac{\sqrt{9-x^2}}{x^2}dx$$ $x=3\sin t,dx=3\cos tdt$ $$\int\frac{\sqrt{9-9\sin^2t}}{9\sin^2t}3\cos tdt=\int\frac{\cos^2t}{\sin^2t} dt\neq3\int\frac{\cos^2t}{\sin^2t} dt$$

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$$\int\frac{\sqrt{9-x^2}}{x^2}dx=\vert x=3\sin t\Rightarrow dx=-3\cos t dt\vert=\int\frac{\sqrt{9-9\sin^2t}}{9\sin^2 t}\cdot (-3\cos t) dt=$$$$\int\frac{\sqrt{9(1-\sin^2 t)}}{9\sin^2 t}\cdot (-3\cos t) dt=\int\frac{-3\sqrt{1-\sin^2 t}}{9\sin^2 t}\cdot 3\cos t dt=$$ $$-\int\frac{\cos^2 t}{\sin^2 t }dt=-\int\cot^2 t dt=-(-\cot t-t)=\cot t+t=$$

You need only to substitution $t$ for $t=\arcsin \frac{x}{3}$

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