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Question
Evaluate the following:
`int ("d"x)/sqrt(16 - 9x^2)`
Solution
Let I = `int ("d"x)/sqrt(16 - 9x^2)`
= `1/3 int ("d"x)/sqrt(16/9 - x^2)`
= `1/3 int ("d"x)/sqrt((4/3)^2 - x^2)`
= `1/3 sin^-1 x/(4/3) + "C"` ....`[because int ("d"x)/sqrt("a"^2 - x^2) = sin^-1 x/"a" + "C"]`
∴ I = `1/3 sin^-1 (3x)/4 + "C"`
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