How to convert sin2∅ in trigonometry substituion.

[clifftan]

the question is ∫ √(y^2-25)/y^3 dy..

I did found the answer which is 1/5 sec^-1(y/5)-sin2∅/20-∅/10
But mine tutorial answer is 1/5 sec^-1(y/5)- √(y^2-25)/2y^2+c
i don't know how to convert the behind part sin2∅/20-∅/10 into √(y^2-25)/2y^2

 

[chiro]

The typical trig substitution is where sin^2 + cos^2 = 1.

Now you should let y^2 - 25 = (25/25)*y^2 - 25 = (25/25)*y^2 - 25 = 25[y^2/25 - 1]
= 25[(y^2/25 - 1].

Have a new variable where z = y/SQRT(25) (through an integral substitution) and you'll have to clean up a little but this term will go to [z^2 - 1] which can use a trig substitution of either sin or cos depending on what the other terms are like.

So to start off for your term involving the square root, factor out the 25 from the square root which will leave you with 5*SQRT(y^2/25 - 1)/2y^2 and make a substitution z = y/SQRT(25) which means dz = dy/5. Now clean up the square root sign to get 5*SQRT(z^2-1)/(2*25*z^2) * 5*dz = SQRT(z^2 - 1)/2z^2.

 

[clifftan]

Thank you very much~~