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Ch.8 Extra Topic: Trigonometric Substitutions Monticello (Thomas Jefferson’s home), Alexandria, VA Photo by Vickie Kelly, 2004 Greg Kelly, Hanford High School, Richland, Washington We can use right triangles and the pythagorean theorem to simplify some problems. 1 dx 4 x 2 These are in the same form. a4 xx ln sec tan C 4 x x C 2 2 2 ln x 2sec2 d 2sec sec d 22 2 a2 4 x2 sec 2 2sec 4 x 2 x tan 2 2 tan x 2sec 2 d dx We can use right triangles and the pythagorean theorem to simplify some problems. 1 dx 4 x2 2sec2 d 2sec 4 x2 x C 2 ln ln 4 x 2 x ln 2 C sec d ln sec tan C ln 4 x2 x C 2 2 This is a constant. ln 4 x2 x C This method is called Trigonometric Substitution. a x 2 a x , 2 If the integral contains 2 2 x we use the triangle at right. a If we need a2 x2 , we move a to the hypotenuse. If we need move x to the hypotenuse. x a x a x 2 x2 a2 , we 2 x2 a2 a 2 x 2 dx 9 x2 3 9 x2 9sin 2 3cos d 3cos 1 cos 2 9 d 2 9 1 cos 2 d 2 9 9 1 sin 2 C 2 2 2 x x sin 3 3sin x 9 x2 cos 3 3cos 9 x2 3cos d dx x sin 3 double angle x sin 1 formula 3 9 1 x 9 sin 2sin cos C 2 3 4 9 1 x 9 x 9 x 2 sin C 2 3 2 3 3 2 x 2 dx 9 x2 3 9 x2 x x sin 3 3sin x 9 x2 cos 3 3cos 9 x2 3cos d dx x sin 3 double angle x sin 1 formula 3 9 1 x x sin 9 x2 C 2 3 2 9 1 x 9 sin 2sin cos C 2 3 4 9 1 x 9 x 9 x 2 sin C 2 3 2 3 3 5 dx 2 x x2 2x x 2 dx 1 x 1 2 Let u x 1 x2 2 x du dx x2 2x 1 1 du 1 u2 cos d cos d We can get 2x x 2 into the necessary form by completing the square. x 1 1 2 1 x 1 2 1 sin u C C sin 1 x 1 C 1 u 1 u2 1 u2 1 u2 cos 1 sin u cos d du 6 dx 4x2 4x 2 4 x2 4 x 2 4 x2 4 x 1 1 dx 2 x 1 Complete the square: 2 1 1 du 2 u2 1 1 sec2 d 2 sec2 2 x 1 2 1 Let u 2x 1 du 2 dx 1 du dx 2 1 1 1 1 tan u C C d 2 2 2 1 tan 1 2 x 1 C 2 u2 1 u 1 tan u sec2 d du sec u 2 1 sec2 u 2 1 Here are a couple of shortcuts that are result from Trigonometric Substitution: du 1 1 u u 2 a 2 a tan a C du u sin C a a2 u 2 1 These are on your list of formulas. They are not really new. p