Integration help please.

[purpledragooon]

integrate dy/((y(y^2-1))^(1/2)=(pi/12) lower limit is (2)^(1/2), upper limit is unkown or x.
i tried to just derive it first and got -tan^-11+tan^-1(1/(x^2-1)^(1/2))=pi/12. from here im lost please help.

 

[Yogi]

Whenever you see a square root of a sum or difference of squares you should always think "trig substitution". Set up a right triangle and convert from y to theta first.

 

[purpledragooon]

Whenever you see a square root of a sum or difference of squares you should always think "trig substitution". Set up a right triangle and convert from y to theta first.

i solved the integral now i need to find the x but i can't figure it out.

 

[Yogi]

i solved the integral now i need to find the x but i can't figure it out.

What did you get?

Show your work/results via trig substitution and we can continue.

 

[soroban]

equation

Hello, purpledragooon!

\(\displaystyle \displaystyle\text{Find the value of }x\text{ that satisfies the equation: }\:\int^x_{\sqrt{2}}\frac{dy}{y\sqrt{y^2-1}}\:=\:\frac{\pi}{12} \)

The integral is a standard form: arcsecant.

We have: .\(\displaystyle \text{arcsec}\,y\,\bigg]^x_{\sqrt{2}} \;=\;\dfrac{\pi}{12} \)

. \(\displaystyle \text{arcsec}\,x - \text{arcsec}\,\sqrt{2} \;=\;\dfrac{\pi}{12}\)

. . . . . . .\(\displaystyle \text{arcsec}\,x - \dfrac{\pi}{4} \;=\;\dfrac{\pi}{12}\)

. . . . . . . . . . \(\displaystyle \text{arcsec}\,x \;=\; \dfrac{\pi}{3}\)

. . . . . . . . . . . . . . .\(\displaystyle x \;=\;\sec\frac{\pi}{3}\)

. . . . . . . . . . . . . . .\(\displaystyle x \;=\;2\)