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.

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.