Using trigonometric substitution to evaluate * ,5x³√9-x² dx 243 243 45 - 15 (1-x²) ³² + C 243 5 ² (1 − x ²) ² — ²3³ (1 − x ² )² + C 243 2 + ³ (₁-²) ² - ² + ³ (₁ - ) ² + 0 243 5 3 none of the choices 5 243 (1-x²)z

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Q: 3 We were given that = 1²ex dx = @³² = e. 3 3 Therefore, •L²³ ex+ 5 dx = 20²³ [²³ ex α 2e5 dx =

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