How can I evaluate the integral using a trigonometric identity?

[karush]

Evaluate using a trig identity
$$\displaystyle
\int \dfrac{5}{x^2\sqrt{25-x^2}}\, dx$$
my first inclination to set
$u=5\sin{x}$
then
$du=5\cos{x}\, dx$ or $dx=\dfrac{du}{5\cos {x}}$

 

[skeeter]

$x=5\sin{t} \implies dx=5\cos{t} \, dt$

$$\int \dfrac{25\cos{t}}{25\sin^2{t}\sqrt{25-25\sin^2{t}}} \, dt$$

$$\dfrac{1}{5} \int \csc^2{t} \, dt = -\dfrac{1}{5}\cot{t}+C$$

$t=\arcsin\left(\dfrac{x}{5}\right) \implies -\dfrac{1}{5}\cot{t}=-\dfrac{\sqrt{25-x^2}}{5x}$

 

[karush]

View attachment 9286

Ok this what it was on my cell phone

 

[skeeter]

Ok this what it was on my cell phone

?

here is a pic ... see if your phone renders it ok

 

[karush]

Yeah thanks

That was weird
..

 

[Monoxdifly]

If I recall, smartphones can't read LaTeX, though that was before 2018. I don't know how it is now.