Question

Evaluate the integrals in Exercises 31–78.71. ∫(from √2/3 to 2/3)dy/(|y|√(9y²-1))

Accepted Answer

First, carefully analyze the integral $$ \int_{\sqrt{2}/3}^{2/3} \frac{dy}{|y| \sqrt{9y^2 - 1}} . $$Notice the absolute value $$ |y|  in $$the denominator and the square root expression $$ \sqrt{9y^2 - 1} . $$Since the limits of integration are positive (both $$ \sqrt{2}/3 $$ and $$ 2/3 $$ are positive), we can simplify $$ |y| = y  in $$this interval. Rewrite the integral as $$ \int_{\sqrt{2}/3}^{2/3} \frac{1}{y \sqrt{9y^2 - 1}} \, dy . To $$simplify the integrand, consider a substitution that will help handle the $$ \sqrt{9y^2 - 1} $$ term. A good choice is to let $$ 3y = \sec(	heta) $$, because $$ \sec^2(	heta) - 1 = 	an^2(	heta) $$, which will simplify the square root. Express $$ y $$ and $$ dy  in $$terms of $$ 	heta $$: $$ y = \frac{\sec(	heta)}{3} $$ and $$ dy = \frac{\sec(	heta) 	an(	heta)}{3} d	heta . $$Also, rewrite the square root: $$ \sqrt{9y^2 - 1} = \sqrt{\sec^2(	heta) - 1} = 	an(	heta) . $$Substitute all these into the integral. After substitution, the integral becomes $$ \int \frac{1}{(\frac{\sec(	heta)}{3}) \cdot 	an(	heta)} \cdot \frac{\sec(	heta) 	an(	heta)}{3} d	heta . $$Simplify the expression inside the integral by canceling terms carefully. Finally, determine the new limits of integration by substituting the original $$ y -$$values into $$ 3y = \sec(	heta)  to $$find the corresponding $$ 	heta -$$values. Then, integrate the simplified function with respect to $$ 	heta $$ over these new limits.

Evaluate the integrals in Exercises 31–78.
71. ∫(from √2/3 to 2/3)dy/(|y|√(9y²-1))

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Key Concepts

Definite Integrals with Absolute Value

Integration of Functions Involving Square Roots

Domain and Limits of Integration