Trig-sub Integration

**cdlegendary** · February 16th 2010, 11:33 AM

Does anyone know how to do the above integral? I don't know what I should sub into the equation

**Krizalid** · February 16th 2010, 11:39 AM

if you want a trig. substitution, then put $x=5\sin t.$

or put $x=\frac1t.$

**cdlegendary** · February 16th 2010, 11:43 AM

Originally Posted by Krizalid

if you want a trig. substitution, then put $x=5\sin t.$

or put $x=\frac1t.$

thanks, but could you explain briefly how one would know to choose those substitutions?

**Soroban** · February 16th 2010, 11:44 AM

Hello, cdlegendary!

How can you not know?

This is the very first trig substitution you ever learned . . .

$\int\frac{-2\,dx}{x^2\sqrt{25-x^2}}$

Let: . $x \:=\:5\sin\theta \quad\Rightarrow\quad dx \:=\:5\cos\theta\,d\theta$

Also: . $\sqrt{25-x^2} \:=\:5\cos\theta$

Substitute: . $-2\int\frac{5\cos\theta\,d\theta}{25\sin^2\!\theta\ cdot 5\cos\theta} \;=\;-\frac{2}{25}\int\frac{d\theta}{\sin^2\!\theta} \;=\;-\frac{2}{25}\int\csc^2\!\theta\,d\theta$

Can you finish it now?

**Krizalid** · February 16th 2010, 11:46 AM

let's make it more generally:

suppose you have an expression in the integrand, say $\sqrt{a^2-x^2},$ so we want to delete the square root in order to make the integration easy, so in this case we invoke the trig. identity $\sin^2t+\cos^2t=1.$ Now if we put $x=a\sin t$ then $a^2-x^2=a^2-a^2\sin^2t=a^2(1-\sin^2t)=a^2\cos^2t,$ so yes, it's just taking the square root of a squared thing.

**cdlegendary** · February 16th 2010, 11:48 AM

Thanks for the help guys, its much appreciated. I can figure it out from here.

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