# Trigonometric substitution for $\int\frac{1}{x^2\sqrt{4-x^2}}dx$

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.

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 $x$.

• I still don't get how you got $$\frac{1}{4}\int\frac{1}{\sin^2\theta}\mathrm{d}\theta=\frac{-\cos\theta}{4\sin\theta}$$ May 14, 2014 at 8:35
• 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)
– john
May 14, 2014 at 8:37
• alternatively, you could equivalently differentiate $\frac{-1}{4\tan\theta}$ using the chain rule and then simplify
– john
May 14, 2014 at 8:39
• 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 $$\frac{1}{4}\int\frac{1}{\sin^2\theta}\mathrm{d}\theta$$ May 14, 2014 at 8:40
• 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 $4$.
– john
May 14, 2014 at 8:46