I was asked to evaluate $$\displaystyle\int\frac{1}{x^2\sqrt{1-x^2}}\text{d}x$$

Here's my attempt:

$$\text{Let x=\sin(\theta)}$$

$$\text{Then }\text{d}x=\cos(\theta)\text{d}\theta$$ $$\int\frac{1}{\sin^2(\theta)\sin(\theta)}\cos(\theta)\text{d}\theta$$ $$\int\cos(\theta)\sin^{-3}(\theta)\text{d}\theta$$ $$\text{Let u = \sin(\theta)}$$

$$\text{Then \text{d}u/\cos(\theta) = \text{d}x}$$ $$\int u^{-3}\text{d}u$$ $$-\frac{1}{2}u^{-2}$$ $$-\frac{1}{2\sin^2(\theta)}$$ $$-\frac{1}{2x^2}+C$$

But the book says:

$$-\frac{\sqrt{1-x^2}}{x}$$

Where's the error in my reasoning? Thanks.

• $$\sqrt{1-\sin^2x}=\cos x$$ up to sign. – DonAntonio Aug 15 at 15:48

You will have $$\cos(\theta)$$ in the denominator at the first step of your trig substitution.

Set $$x=\sin t,-\dfrac\pi2\le t\le\dfrac\pi2$$

$$\cos t=+\sqrt{1-x^2}$$

$$dx=?$$ to find

$$\int\dfrac{\cos t\ dt}{\sin^2t\cos t}=-\cot t+K$$

it it $$\sqrt{1-\sin^2(\theta)}=\pm\cos(\theta)$$

To add to Lab Bhattacharjee's hint: draw a triangle with sides $$1$$ and $$x$$, find the third side with the Pythagorean theorem, and then remember that $$-\cot x = -\dfrac {\cos x}{\sin x}$$.