# 8.2.7-eg - y = 0 ⇒ u = 1 y = 1 2 ⇒ u = 0 to obtain Z 1...

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Example Evaluate Z 1 / 2 0 sin - 1 (2 y ) dy . If we let u = sin - 1 (2 y ) , dv = dy, du = 2 1 - 4 y 2 dy, v = y, then the integral is of the form R b a u dv , so we use the integration by parts formula Z b a u dv = uv ± ± ± ± ± b a - Z b a v du. Therefore Z 1 / 2 0 sin - 1 (2 y ) dy = y sin - 1 (2 y ) ± ± ± ± ± 1 / 2 0 - Z 1 / 2 0 2 y dy 1 - 4 y 2 = 1 2 sin - 1 (1) - 0 - Z 1 / 2 0 2 y dy 1 - 4 y 2 = π 4 - Z 1 / 2 0 2 y dy 1 - 4 y 2 (1) It remains to evaluate Z 1 / 2 0 2 y dy 1 - 4 y 2 . Make the substitution u = 1 - 4 y 2 , du = - 8 y dy and adjust the limits of integration
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Unformatted text preview: y = 0 ⇒ u = 1 y = 1 / 2 ⇒ u = 0 , to obtain Z 1 / 2 2 y dy √ 1-4 y 2 = 1 4 Z 1 u-1 / 2 du = 1 2 u 1 / 2 ± ± ± ± ± 1 = 1 2 . (2) Substituting (2) into (1), we have Z 1 / 2 sin-1 y dy = π 4-1 2 ....
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