Ex 7.4, 25 - Chapter 7 Class 12 Integrals

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Ex 7.4, 25 ∫1▒𝑑𝑥/√(9𝑥 − 4𝑥^2 ) equals A. 1/9 sin−1 ((9𝑥 − 8)/8) + C B. 1/9 sin−1 ((8𝑥 − 9)/9) + C C. 1/3 sin−1 ((9𝑥 − 8)/8) + C D. 1/2 sin−1 ((9𝑥 − 8)/8) + C ∫1▒𝑑𝑥/√(9𝑥 − 4𝑥^2 ) =∫1▒𝑑𝑥/√(−4(𝑥^2 − 9/4 𝑥) ) =∫1▒𝑑𝑥/√(−4(𝑥^2 − 2(𝑥) (9/8)) ) (Taking −4 common) =∫1▒𝑑𝑥/√(−4[𝑥^2 − 2(𝑥) (9/8) + (9/8)^2− (9/8)^2 ] ) =∫1▒𝑑𝑥/√(−4[(𝑥 − 9/8)^2− (9/8)^2 ] ) =∫1▒𝑑𝑥/√(4[(9/8)^2 − (𝑥 − 9/8)^2 ] ) =∫1▒𝑑𝑥/(√4 √((9/8)^2 − (𝑥 − 9/8)^2 )) =1/2 ∫1▒𝑑𝑥/√((9/8)^2 − (𝑥 − 9/8)^2 ) It is of form ∫1▒𝑑𝑥/√(𝑎^2 − 𝑥^2 ) =sin^(−1)⁡〖𝑥/𝑎〗 +𝐶1 ∴ Replacing 𝑥 by (𝑥− 9/8) and 𝑎 by 9/8 , we get =1/2 [sin^(−1)⁡〖(𝑥 − 9/8)/(9/8)〗 +𝐶1] =1/2 sin^(−1)⁡[(𝑥 − 9/8)/(9/8)] +𝐶 =1/2 sin^(−1)⁡〖((8𝑥 − 9)/8)/(9/8)〗 +𝐶 =1/2 sin^(−1)⁡〖(8𝑥 − 9)/9〗 +𝐶 ∴ Option B is correct answer