Does anyone know how to do the above integral? I don't know what I should sub into the equation
Hello, cdlegendary!
How can you not know?
This is the very first trig substitution you ever learned . . .
$\displaystyle \int\frac{-2\,dx}{x^2\sqrt{25-x^2}} $
Let: .$\displaystyle x \:=\:5\sin\theta \quad\Rightarrow\quad dx \:=\:5\cos\theta\,d\theta$
Also: .$\displaystyle \sqrt{25-x^2} \:=\:5\cos\theta$
Substitute: .$\displaystyle -2\int\frac{5\cos\theta\,d\theta}{25\sin^2\!\theta\ cdot 5\cos\theta} \;=\;-\frac{2}{25}\int\frac{d\theta}{\sin^2\!\theta} \;=\;-\frac{2}{25}\int\csc^2\!\theta\,d\theta$
Can you finish it now?
let's make it more generally:
suppose you have an expression in the integrand, say $\displaystyle \sqrt{a^2-x^2},$ so we want to delete the square root in order to make the integration easy, so in this case we invoke the trig. identity $\displaystyle \sin^2t+\cos^2t=1.$ Now if we put $\displaystyle x=a\sin t$ then $\displaystyle a^2-x^2=a^2-a^2\sin^2t=a^2(1-\sin^2t)=a^2\cos^2t,$ so yes, it's just taking the square root of a squared thing.