Mathematics

Question

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

1 Answer

  • [tex]\displaystyle\int\frac{\mathrm dx}{\sqrt{81-225x^2}}[/tex]

    Let [tex]x=\dfrac9{15}\sin t[/tex], so that [tex]\mathrm dx=\dfrac9{15}\cos t\,\mathrm dt[/tex] and the integral becomes

    [tex]\displaystyle\int\frac{\dfrac9{15}\cos t}{\sqrt{81-225\left(\frac9{15}\sin t\right)^2}}\,\mathrm dt[/tex]
    [tex]\displaystyle\int\frac{\dfrac9{15}\cos t}{\sqrt{81-225\times\frac{81}{225}\sin^2 t}}\,\mathrm dt[/tex]
    [tex]\displaystyle\frac1{15}\int\frac{\cos t}{\sqrt{1-\sin^2 t}}\,\mathrm dt[/tex]
    [tex]\displaystyle\frac1{15}\int\frac{\cos t}{\sqrt{\cos^2 t}}\,\mathrm dt[/tex]
    [tex]\displaystyle\frac1{15}\int\frac{\cos t}{|\cos t|}\,\mathrm dt[/tex]

    When [tex]\cos t>0[/tex], you have [tex]|\cos t|=\cos t[/tex]. Under these conditions, you can write

    [tex]\displaystyle\frac1{15}\int\frac{\cos t}{\cos t}\,\mathrm dt=\frac1{15}\int\mathrm dt=\frac t{15}+C[/tex]

    and back-substituting yields

    [tex]\dfrac{\arcsin\frac{15x}9}{15}+C=\dfrac{\arcsin\frac{5x}3}{15}+C[/tex]
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