Section7_5_review

# Section7_5_review - . Solution. sin x 1 + cos 2 x dx = ?...

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Section 7.5: Inverse Trigonometric Functions The inverse tangent function fx (29 = tan - 1 x exists for fx (29 = tan x , - π 2 < x < 2 . Note that the domain of fx (29 = tan - 1 x is -∞ , ( 29 and the range is - 2 , 2 . The inverse sine function fx (29 = sin - 1 x exists for f x (29 = sin x , - 2 x 2 . Note that the domain of fx (29 = sin - 1 x is - 1 ,1 [ ] and the range is - 2 , 2 . The inverse cosine function fx (29 = cos - 1 x exists for fx ( 29 = cos x , 0 x ≤π . Note that the domain of fx (29 = cos - 1 x is - 1 ,1 [ ] and the range is 0, [ ] . Differentiation/Integration formulas involving the inverse trig. functions: d dx tan - 1 u ( 29 = 1 1 + u 2 du dx d dx sin - 1 u ( 29 = 1 1 - u 2 du dx d dx cos - 1 x ( 29 =- 1 1 - u 2 du dx 1 x 2 + 1 dx = tan - 1 x + c 1 x 2 + a 2 dx = 1 a tan - 1 x a + c 1 1 - x 2 dx = sin - 1 x + c Problem. Differentiate y = cos - 1 3 x 2 ( 29 . Solution. d dx cos - 1 u = - 1 1 - u 2 du dx y = - 1 1 - 3 x 2 ( 29 2 d dx 3 x 2 ( 29 = - 6 x 1 - 9 x 4 Problem. Differentiate f t ( 29 = tan - 1 1 - t . Solution. d dx tan - 1 u = 1 1 + u 2 du dx df dt = 1 1 + 1 - t ( 29 2 d dt 1 - t = 1 1 + 1 - t 1 2 1 - t ( 29 - 1 2 - 1 ( 29 = - 1 2 2 - t ( 29 1 - t

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Problem. Differentiate y = sin - 1 2 x + 1 ( 29 . Solution. d dx sin - 1 u = 1 1 - u 2 du dx dy dx = 1 1 - 2 x + 1 ( 29 2 d dx 2 x + 1 ( 29 = 2 1 - 2 x + 1 ( 29 2 Problem. Determine tan - 1 x 1 + x 2 dx . Solution . tan - 1 x 1 + x 2 dx = ? Let u = tan - 1 x , then du = 1 1 + x 2 dx . tan - 1 x 1 + x 2 dx = udu = u 2 2 + c = 1 2 tan - 1 x ( 29 2 + c Problem. Evaluate sin x 1 + cos 2 x dx
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Unformatted text preview: . Solution. sin x 1 + cos 2 x dx = ? Let u = cos x , then du = -sin xdx and sin xdx = -du . sin x 1 + cos 2 x dx = 1 1 + u 2-du ( 29 = -1 1 + u 2 du = -tan-1 u + c = -tan-1 cos x ( 29 + c Problem. Evaluate dx 2 + x-1 ( 29 2 . Solution. dx 2 + x-1 ( 29 2 = ? Let u = x-1 , then du = dx . dx 2 + x-1 ( 29 2 = du 2 + u 2 = 1 2 tan-1 u 2 + c = 1 2 tan-1 x-1 2 + c Problem. Evaluate e sin-1 x x-x 2 dx . Solution. e sin-1 x x-x 2 dx = ? Let u = sin-1 x , then du = 1 1-x ( 29 2 1 2 x-1 2 dx du = 1 1-x 1 2 x dx = 1 2 x 1-x dx = 1 2 x 1-x ( 29 dx Therefore, 2 du = 1 2 x-x 2 dx . e sin-1 x x-x 2 dx = e u 2 du = 2 e u + c = 2 e sin-1 x + c...
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## This note was uploaded on 01/12/2010 for the course MATH 44245 taught by Professor Famiglietti during the Fall '07 term at UC Irvine.

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Section7_5_review - . Solution. sin x 1 + cos 2 x dx = ?...

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