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Does anyone know how to do trigonometric substitutions by letting u=x and a=#, and solving with the triangle? If (25x^2 - 36) is under the radical, what do you use for x and the #?
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Sure. First factor stuff out. \[\sqrt{25x^2-36}= \sqrt{36(25x^2/36)-1)} = 6\sqrt{(5x/6)-1}\]
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