Question 8 β« ππ₯/β(9 β 25π₯^2 ) (a) sin-1 (5π₯/3)+ c (b) 1/5 sin-1 (5π₯/3) + c (c) 1/6 log ((3 + 5π₯)/(3 β5π₯)) + c (d) 1/30 log ((3 + 5π₯)/(3 β5π₯)) + c
β«1β1/β(9 β 25π₯^2 ) ππ₯
=β«1β1/β(25(9/25 β π₯^2 ) ) ππ₯
=β«1β1/(β25 β(9/25 β π₯^2 )) ππ₯
=1/5 β«1β1/β(9/25 β π₯^2 ) ππ₯
=1/5 β«1β1/β((3/5)^2 β π₯^2 ) ππ₯
It is of the form
β«1βππ₯/β(π^2 β π₯^2 ) =γπ ππγ^(β1)β‘γπ₯/πγ +π
β΄ Replacing π by 3/5 , we get
=1/5 [sin^(β1)β‘γπ₯/(3/5)γ +πΆ1]
=π/π γπππγ^(βπ)β‘γππ/πγ +πͺ
So, (b) is the correct answer

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.