INTEGRATION

Problem : 13
Integrate ∫ [24 dy / y √(y² - 16)]


Answer :
Let I = ∫ [ 24  dy / y√(y² - 16) ]
=> 24 ∫ [ y dy / y²√(y² - 16) ]


put y² - 16 = t² => y dy = t dt

=> 24 ∫ [ t dt / (t² + 16)√(t²) ]
=> 24 ∫ [ t dt / (t² + 16) * t ]
=> 24 ∫ [ dt / (t² + 16) ]
=> 24 ∫ [ dt / (t² + 4²) ]
=> 24 * (1/4) tan^(-1) (t/4) + C

=> 6 * tan^(-1) [√(y² - 16) / 4] + C (Answer)
OR
 6 * cos^(-1) ( 4/y ) + C (Answer)
OR
6 * sec^(-1) ( y/4) + C (Answer)