1. ## Trig-sub Integration

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

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

or put $x=\frac1t.$

3. 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?

4. 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?

5. 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.

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