print barcode printer c# = x 2 sin x + 2x cos x 2 sin x + C in Software

Painting QR Code 2d barcode in Software = x 2 sin x + 2x cos x 2 sin x + C

= x 2 sin x + 2x cos x 2 sin x + C
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Solutions to Exercises
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Notice that t sin 3t cos 3t dt = 1 t sin 6t dt Now let u = t and 2 dv = sin 6t dt Then 1 1 1 1 t sin 6t dt = cos 6t 1 dt t cos 6t 6 6 2 2 t 1 = cos 6t + sin 6t + C 72 12 2 We do (a), (b), (c), (d) (a) 1/7 1/7 1 = + hence (x + 2)(x 5) x+2 x 5 dx = (x + 2)(x 5) = 1/7 + x+2 1/7 x 5
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1 1 ln |x + 2| + ln |x 5| + C 7 7 1/2 x/2 + 1/2 1 = hence + 2 + 1) x+1 x2 + 1 (x + 1)(x dx = (x +1)(x 2 +1) 1/2 dx + x +1 x/2 dx + x 2 +1 1/2 dx x 2 +1
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1 1 1 = ln |x +1| ln |x 2 +1|+ Tan 1 x +C 4 2 2 (c) Now x 3 2x 2 5x + 6 = (x 3)(x + 2)(x 1) Then x3 As a result, dx x 3 2x 2 5x +6 = = (d) 1/10 dx + x 3 1/15 dx + x +2 1/6 dx x 1 2x 2 1/10 1/15 1/6 1 = + + x 3 x+2 x 1 5x + 6
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1 1 1 ln |x 3|+ ln |x +2| ln |x 1|+C 10 15 6
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Now x 4 1 = (x 2 1)(x 2 + 1) = (x 1)(x + 1)(x 2 + 1) Hence 1/4 x 1/4 x/2 = + + 2 4 1 x 1 x+1 x +1 x We conclude that 1 x dx 1 1 = ln |x 1| + ln |x + 1| ln |x 2 + 1| + C 4+1 4 4 4 x
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3 We do (a), (b), (c), (d) (a) Let u = sin x, du = cos x dx Then the integral becomes (1 + u2 )3 + C 3 Resubstituting x, we obtain the nal answer (1 + u2 )2 2u du = (1 + sin2 x)2 2 sin x cos x dx = (b) Let u = (1 + sin2 x)3 + C 3
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x, du = 1/[2 x] dx Then the integral becomes 2 sin u du = 2 cos u + C
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Resubstituting x, we obtain the nal answer sin x dx = 2 cos x + C x (c) Let u = ln x, du = [1/x] dx Then the integral becomes 1 1 sin 2u du = cos 2u + C 4 2 Resubstituting x, we obtain the nal answer cos u sin u du = 1 cos(ln x) sin(ln x) dx = cos(2 ln x) + C x 4 (d) Let u = tan x, du = sec2 x dx Then the integral becomes eu du = eu + C Resubstituting x, we obtain the nal answer etan x sec2 x dx = etan x + C 4 We do (a), (b), (c), (d) (a) Let u = cos x, du = sin x dx Then the integral becomes u3 + C 3 Resubstituting x, we obtain the nal answer u2 du = sin x cos2 x dx = cos3 x + C 3
(b) Write sin3 x cos2 x dx =
Solutions to Exercises
sin x(1 cos2 x) cos2 x dx
Let u = cos x, du = sin x dx Then the integral becomes (1 u2 )u2 du = u3 u5 + + C 3 5
Resubstituting x, we obtain the nal answer sin3 x cos2 x dx = (c) cos5 x cos3 x + + C 3 5
Let u = tan x, du = sec2 x dx Then the integral becomes u3 du = u4 + C 4
Resubstituting x, we obtain the nal answer tan3 x sec2 x dx = (d) tan4 x + C 4
Let u = sec x, du = sec x tan x Then the integral becomes u2 du = u3 + C 3
Resubstituting x, we obtain the nal answer tan x sec3 x dx = 5 We do (a), (b), (c), (d) (a) Use integration by parts twice:
1 0 1
sec3 x + C 3
ex sin x dx = sin x ex
ex cosx dx
1 x 0
= [e sin 1 0] cosxe = e sin 1 e cos1+1
0 1 0
ex ( sin x)dx
ex sin x dx
7
We may now solve for the desired integral:
ex sin x dx =
1 [e sin 1 e cos 1] 2
Integrate by parts with u = ln x, dv = x 2 dx Thus
x 2 ln x dx = ln x =1 e3 e3 3
x3 3
x3 1 dx 3 x x3 6
0 e3
13 3
13 + = 3 9 9 (c) We write x 2 (x Thus
1 1 1 2x + 1 = + 2+ x x+1 x + 1)
4 2 4 2 4 2
(2x + 1) dx = x3 + x2
1 dx + x
1 dx + x2
1 dx x+1
= [ln 4 ln 2] + = ln (d) We write
1 1 + [ln 3 ln 5] 4 2
6 1 + 5 4
0 0
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