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I'm reviewing my quizzes to study for midterm tomorrow, and I came across a problem where I'm supposed to integrate:
$$\int\frac{1}{x^2\sqrt{4-x^2}}dx$$

I used Mathematica to solve the problem and I'm sure it gave me the correct answer, which is: $$-\frac{\sqrt{4-x^2}}{4x}$$

I used $ x = 2\sin{\theta}$ and $dx = 2\cos{\theta}$ $d\theta$ to solve the problem, and I only got to $$\frac{1}{8}\int{\frac{1}{\sin^2{\theta}\cos{\theta}}}d\theta$$ Looking at step-by-step solution via WolframAlpha, they used $\theta = \arcsin{\frac{x}{2}}$ to solve the problem which I do not know how to. I don't think there is a need for $\theta = \arcsin{\frac{x}{2}}$ to solve the problem, and I'm wondering if anyone can show me how to solve this step by step without the use of $\theta = \arcsin{\frac{x}{2}}$? Maybe help me understand how to?

Trigonometric substitution is the only method that I'm struggling with, and any tips on improving trig sub skill would be appreciated too.

Thanks.

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Note that if $\theta=\arcsin\frac{x}{2}$ then $x=2\sin\theta$, so they're using the same substitution as you.

When you include your point that $\mathrm{d}x =2\cos\theta\mathrm{d}\theta$, you're looking for $$\frac{1}{4}\int\frac{1}{\sin^2\theta}\mathrm{d}\theta=\frac{-\cos\theta}{4\sin\theta}$$ You should confirm this by differentiating the right hand side.

Now try to draw a triangle which would give rise to the relationship $x=2\sin\theta\,$ in order to obtain an expression in terms of $x$.

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  • $\begingroup$ I still don't get how you got $$\frac{1}{4}\int\frac{1}{\sin^2\theta}\mathrm{d}\theta=\frac{-\cos\theta}{4\sin\theta}$$ $\endgroup$ – Kibbles May 14 '14 at 8:35
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    $\begingroup$ Try using the quotient rule on the right hand side. Otherwise you could recognise this as $\int \mathrm{cosec}^2x \mathrm{d}x=-\cot x$ (but probably not) $\endgroup$ – john May 14 '14 at 8:37
  • $\begingroup$ alternatively, you could equivalently differentiate $\frac{-1}{4\tan\theta}$ using the chain rule and then simplify $\endgroup$ – john May 14 '14 at 8:39
  • $\begingroup$ Oops, I think I asked a wrong question. I know that $$\frac{1}{4}\int\frac{1}{\sin^2\theta}\mathrm{d}\theta=\frac{-\cos\theta}{4\sin\theta}$$, but I don't know where you got $$\frac{1}{4}\int\frac{1}{\sin^2\theta}\mathrm{d}\theta$$ $\endgroup$ – Kibbles May 14 '14 at 8:40
  • $\begingroup$ Oh. In your post, I believe you forgot to substitute that $\mathrm{d}x =2\cos\theta\mathrm{d}\theta$. This extra factor removes the $\cos\theta$ from the denominator as well as changing the $8$ to a $4$. $\endgroup$ – john May 14 '14 at 8:46

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