0
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ $$\sqrt{1-\sin^2x}=\cos x$$ up to sign. $\endgroup$
    – DonAntonio
    Aug 15, 2019 at 15:48

4 Answers 4

5
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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

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

$dx=?$ to find

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $-\cot x = -\dfrac {\cos x}{\sin x}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .