# Calc08_4day2 - 8.4 day 2 Trigonometric Substitutions Greg...

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8.4 day 2 Trigonometric Substitutions Greg Kelly, Hanford High School, Richland, Washington

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We can use right triangles and the pythagorean theorem to simplify some problems. θ a x 2 2 a x + 1 2 4 dx x + These are in the same form. 2 2 4 x + 2 4 sec 2 x θ + = 2 2sec 4 x θ = + tan 2 x θ = 2tan x θ = 2 2sec d dx θ θ = 2 2sec 2sec d θ θ θ sec d θ θ ln sec tan C θ θ + + 2 4 ln 2 2 x x C + + +
We can use right triangles and the pythagorean theorem to simplify some problems. 1 2 4 dx x + 2 2sec 2sec d θ θ θ sec d θ θ ln sec tan C θ θ + + 2 4 ln 2 2 x x C + + + 2 4 ln 2 x x C + + + 2 ln 4 ln 2 x x C + + - + This is a constant. 2 ln 4 x x C + + +

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a θ x 2 2 a x - This method is called Trigonometric Substitution. If the integral contains , we use the triangle at right. 2 2 a x + If we need , we move a to the hypotenuse. 2 2 a x - If we need , we move x to the hypotenuse. 2 2 x a - θ a x 2 2 a x + θ a x 2 2 x a -
2 2 2 9 x dx x - 3 θ x 2 9 x - sin 3 x θ = 3sin x θ = 3cos d dx θ θ = 2 9 cos 3 x θ - = 2 3cos 9 x θ = - 2 9sin 3cos 3cos d

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