integration no.4

**afeasfaerw23231233** · October 27th 2007, 12:10 AM

revise: posting a wrong question

**galactus** · October 27th 2007, 04:19 AM

Just use trig sub.

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

Let $x=sin{\theta}, \;\ dx=cos{\theta}d{\theta}$

Make the subs and it simplifies down to:

$\int{cot^{2}{\theta}}d{\theta}$

$\int{cot^{n}u}du=\frac{-1}{n-1}cot^{n-1}u-\int{cot^{n-2}u}du$

After you integrate cot, then make the sub ${\theta}=sin^{-1}(x)$ to get it back in terms of x.

You're finished.

**qyzren** · October 27th 2007, 04:35 AM

simple answer: coz you did it wrong.
split it up first: ∫√[1/x^4 + 1/x²] dx
then do the substitution, u = 1/x, dx = -1/u² du

-∫√[u^4 + u²] * 1/u² du

factor out u from inside the square root and cancel one from denominator, you get
-∫√[u²-1]/u du
you can use a trig substitution to solve this.

**afeasfaerw23231233** · October 27th 2007, 08:32 AM

revise: posting a wrong question

**Krizalid** · October 27th 2007, 08:55 AM

Does this make sense?

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