1
$\begingroup$

I have been given the integral in the top-left of the image.

I am fairly new to integration (I know basic integration with powers, how to use integration by substitution and what some trigonometric functions differentiate to) and got stuck trying to simplify this integral (I got up to the integral at the bottom).

Any hints would be great!

enter image description here

$\endgroup$
4
$\begingroup$

$x=\csc t,dx=-\csc t\cot t\ dt$

$-\dfrac\pi2<t<\dfrac\pi2,\cot t=\sqrt{x^2-1}$

$$\int\dfrac{dx}{x^2\sqrt{x^2-1}}=-\int\dfrac{\csc t\cot t}{\csc^2t\cot t}dt=-\int\sin t \ dt=K+\cos t$$

$$\cos t=\dfrac{\cot t}{\csc t}=?$$

$\endgroup$
  • $\begingroup$ Ah thanks, the bit I was missing was writing dx=-csc(t)cot(t). $\endgroup$ – Jamminermit Jul 12 at 16:35
  • $\begingroup$ @Jamminermit, Actually I missed to put $$dt$$ $\endgroup$ – lab bhattacharjee Jul 12 at 18:40
  • $\begingroup$ That’s fine, I knew what you meant. $\endgroup$ – Jamminermit Jul 12 at 19:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.