Question

Evaluate 
`int  1/(16 - 9x^2) dx`

Answer 1

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)`