Question

chegg evaluate the following integral using trigonometric substitution. dx/square root(x^2-81), x > 9

Answer 1

Integration will be ln | ( x /9 ) + ( $\sqrt{(x/9)^{2} - 1 }$ ) | + c  .

Given  ∫ 1 dx / ( $x^{2}$ - 81 )

Put ,

x = 9 secθ

dx = 9 secθ tanθ dθ

and

$\sqrt{x^{2} - 9^{2} }$ = 9 tanθ

Substituting values in  ∫ 1 dx / ( $x^{2}$ - 81 ) ,

∫ (9 secθ tanθ ) dθ / ( 9 tanθ )

∫ secθ dθ = ln | secθ + tanθ | + c

                = ln | ( x /9 ) + ( $\sqrt{x^{2} - 9^{2} }$ / 9 ) | + c

              = ln | ( x /9 ) + ( $\sqrt{(x/9)^{2} - 1 }$ ) | + c

Hence , the integration will be ln | ( x /9 ) + ( $\sqrt{(x/9)^{2} - 1 }$ ) | + c .

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