Question

Integral of 1/sqrt (81-225x^2)

Answer 1

`\displaystyle\int(\mathrm dx)/(√(81-225x^2))`

Let
`x=\frac9{15}\sin t`, so that
`\mathrm dx=\frac9{15}\cos t\,\mathrm dt` and the integral becomes


`\displaystyle\int\frac{\frac9{15}\cos t}{\sqrt{81-225\left(\frac9{15}\sin t\right)^2}}\,\mathrm dt`

`\displaystyle\int\frac{\frac9{15}\cos t}{\sqrt{81-225*(81)/(225)\sin^2 t}}\,\mathrm dt`

`\displaystyle\frac1{15}\int(\cos t)/(√(1-\sin^2 t))\,\mathrm dt`

`\displaystyle\frac1{15}\int(\cos t)/(√(\cos^2 t))\,\mathrm dt`

`\displaystyle\frac1{15}\int(\cos t)/(|\cos t|)\,\mathrm dt`

When
`\cos t>0`, you have
`|\cos t|=\cos t`. Under these conditions, you can write


`\displaystyle\frac1{15}\int(\cos t)/(\cos t)\,\mathrm dt=\frac1{15}\int\mathrm dt=\frac t{15}+C`

and back-substituting yields


`\frac{\arcsin\frac{15x}9}{15}+C=\frac{\arcsin\frac{5x}3}{15}+C`