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Question
Evaluate
`int 1/(16 - 9x^2) dx`
Solution
I = `int 1/(16 - 9x^2) dx`
I = `int 1/(sqrt(16(1 - (9x^2)/16))) dx`
I = `int 1/(4 sqrt(1 - (3x/4)^2)) dx`
∴ t = `3x/4`
dt = `3/4` dx
`4/3 dt = dx`
I = `1/(cancel(4) sqrt(1 - t^2)) cancel(4)/3 dt`.
I = `1/3 int 1/sqrt(1 - t^2) dt`
∴ t = sin u
dt = cos u du
I = `1/3 int 1/sqrt(1 - sin^2 u) . cos u du`
I = `1/3 int 1/sqrt(cos^(2) u). cos u du`.........(sin2u + cos2u = 1. ∴cos2u = 1 - sin2u)
I = `1/3 int (cancel(cos u)/cancel(cos u)) du.`
I = `1/3 int 1 du`
I = `1/3 u`
I = `1/3 sin t` ......(∴ t = sin u u = sin t)
I = `1/3 sin ((3x)/4) + C .....(∴ t = 3x/4)`
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