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Question
Find the integrals of the following:
`1/sqrt(9 + 8x - x^2)`
Solution
`int 1/sqrt(9 + 8x - x^2) "d"x = int ("d"x)/sqrt(9 - (x^2 - 8x))`
= `int ("d"x)/sqrt(9 - [(x - 4)^2 - 4^2]`
= `int ("d"x)/sqrt(9 - [(x - 4)^2 - 16]`
= `int ("d"x)/sqrt(9 - (x - 4)^2 + 16)`
= `int ("d"x)/sqrt(25 - (x - 4)^2`
= `int ("d"x)/sqrt(5^2 - (x - 4)^2`
Put x – 4 = t
dx = dt
`int ("d")/sqrt(9 + 8x - x^2) = int ("d"x)/sqrt(5^2 - "t"^2)`
`int ("d"x)/sqrt("a"^2 - x^2) = sin^-1 (x/"a") + "c"`
`int ("d"x)/sqrt(9 + 8x - x^2) = sin^-1 ("t"/5) + "c"`
= `sin^-1 ((x - 4)/5) + "c"`
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