Integration by specific formulaes - Formula 5

Chapter 7 Class 12 Integrals
Concept wise

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Ex 7.4, 25 β«1βππ₯/β(9π₯ β 4π₯^2 ) equals A. 1/9 sinβ1 ((9π₯ β 8)/8) + C B. 1/9 sinβ1 ((8π₯ β 9)/9) + C C. 1/3 sinβ1 ((9π₯ β 8)/8) + C D. 1/2 sinβ1 ((9π₯ β 8)/8) + C β«1βππ₯/β(9π₯ β 4π₯^2 ) =β«1βππ₯/β(β4(π₯^2 β 9/4 π₯) ) =β«1βππ₯/β(β4(π₯^2 β 2(π₯) (9/8)) ) (Taking β4 common) =β«1βππ₯/β(β4[π₯^2 β 2(π₯) (9/8) + (9/8)^2β (9/8)^2 ] ) =β«1βππ₯/β(β4[(π₯ β 9/8)^2β (9/8)^2 ] ) =β«1βππ₯/β(4[(9/8)^2 β (π₯ β 9/8)^2 ] ) =β«1βππ₯/(β4 β((9/8)^2 β (π₯ β 9/8)^2 )) =1/2 β«1βππ₯/β((9/8)^2 β (π₯ β 9/8)^2 ) It is of form β«1βππ₯/β(π^2 β π₯^2 ) =sin^(β1)β‘γπ₯/πγ +πΆ1 β΄ Replacing π₯ by (π₯β 9/8) and π by 9/8 , we get =1/2 [sin^(β1)β‘γ(π₯ β 9/8)/(9/8)γ +πΆ1] =1/2 sin^(β1)β‘[(π₯ β 9/8)/(9/8)] +πΆ =1/2 sin^(β1)β‘γ((8π₯ β 9)/8)/(9/8)γ +πΆ =1/2 sin^(β1)β‘γ(8π₯ β 9)/9γ +πΆ β΄ Option B is correct answer