Assignment 6 Solution Please do not copy and paste my answer. You will get similar questions but with different numbers! This question tests you the following points: Integration by Parts: Let u = x, dv = cos(2x), then du = dx and v = dv = cos(2x)dx = 1 2 sin(2x). x cos(2x)dx = = x 1 2 sin(2x) 1 2 sin(2x)dx = 1 2 sin(2x)x + 1 cos(2x) +C 4 1
Integration by Parts: Let u = x 2 + 2x, dv = cos(x), then du = (2x + 2)dx = 2(x + 1) and v = dv = cos(x)dx = sin(x). x cos(2x)dx = = (x 2 + 2x)sin(x) 2 (x + 1)sin(x)dx Apply Integration by parts again to find (x + 1)sin(x)dx. Let u = x + 1 and dv = sin xdx, then du = dx and v = dv = sin xdx = cos x (x + 1)sin(x)dx = = (x + 1)cos x + = (x + 1)cos x + sin x cos xdx Plug back to the first step x cos(2x)dx = (x 2 + 2x)sin(x) 2 (x + 1)sin(x)dx = (x 2 + 2x)sin(x) 2( (x + 1)cos x + sin x) +C 2
Integration by Parts: Let u = tan 1 (8t), dv = dt, then du = 8 1+(8t) 2 dt = 8 1+64t 2 dt and v = dv = dt = t. tan 1 (8t)dt = = tan 1 (8t)t 8 t 1 + 64t 2 dt To find t dt, use u-substitution. Let u = 1 + 64t 2, then du = 128tdt or tdt = 1 1+64t 2 128 du Plug back to the previous step: t 1 + 64t 2 dt = 1 1 128 u du = 1 128 ln u = 1 128 ln 1 + 64t 2 = 1 128 ln(1 + 64t 2 ) tan 1 (8t)dt = tan 1 t (8t)t 8 1 + 64t 2 dt = tan 1 (8t)t 8 1 128 ln(1 + 64t 2 ) +C = tan 1 (8t)t 1 16 ln(1 + 64t 2 ) +C 3
Integration by Parts: Let u = ln p, dv = p 3 dp, then du = 1 p dp and v = dv = p 3 dp = 1 4 p4. p 3 ln pdp = 1 1 4 p4 p dp = ln p 1 4 p4 = p4 ln p 4 = p4 ln p 4 1 4 p 3 dp 1 16 p4 +C 4
Integration by Parts: Let u = sin(9θ), dv = e 8θ dθ, then du = 9cos(9θ)dθ and v = dv = e 8θ dθ = 1 8 e8θ. sin(9θ)e 8θ dθ = 1 8 e8θ 9cos(9θ)dθ e 8θ cos(9θ)dθ = sin(9θ) 1 8 e8θ = 1 8 sin(9θ)e8θ 9 8 To find e 8θ cos(9θ)dθ, apply integration by parts again. Let u = cos(9θ) and dv = e 8θ dθ, then du = 9sin(9θ)dθ and v = dv = e 8θ dθ = 1 8 e8θ. inte 8θ cos(9θ)dθ = 1 8 e8θ 9sin(9θ)dθ e 8θ sin(9θ)dθ = cos(9θ) 1 8 e8θ + = 1 8 cos(9θ)e8θ + 9 8 Now plug back to previous step: sin(9θ)e 8θ dθ = 1 8 sin(9θ)e8θ 9 8 e 8θ cos(9θ)dθ = 1 8 sin(9θ)e8θ 9 8 [ 1 8 cos(9θ)e8θ + 9 8 = 1 8 sin(9θ)e8θ 9 64 cos(9θ)e8θ 81 64 e 8θ sin(9θ)dθ] e 8θ sin(9θ)dθ Then sin(9θ)e 8θ dθ + 81 64 145 64 e 8θ sin(9θ)dθ = 1 8 sin(9θ)e8θ 9 64 cos(9θ)e8θ e 8θ sin(9θ)dθ = 1 8 sin(9θ)e8θ 9 64 cos(9θ)e8θ e 8θ sin(9θ)dθ = 64 145 [ 1 8 sin(9θ)e8θ 9 64 cos(9θ)e8θ ] +C 5
Integration by Parts: Let u = z 3, dv = e z dz, then du = 3z 2 dz and v = dv = e z dz = e z. z 3 e z = e z z 2 dz (a) = z 3 e z 3 To find e z z 2 dz, let u = z 2 and dv = e z dz, then du = 2z and v = e z. z 2 e z = e z zdz = z 2 e z 2 (b) To find e z zdz, let u = z and dv = e z dz, then du = z and v = e z. ze z = = ze z e z dz = ze z e z (c) Plug (c) into (b): z 2 e z = z 2 e z 2 e z zdz (b) = z 2 e z 2(ze z e z ) = z 2 e z 2ze z 2e z Then plug it back to (a) z 3 e z = z 3 e z 3 e z z 2 dz (a) = z 3 e z 3(z 2 e z 2ze z 2e z ) = z 3 e z 3z 2 e z + 6ze z 6e z So 19 z 3 e z = 19(z 3 e z 3z 2 e z + 6ze z 6e z ) 6
Integration by Parts: Let u = lnr, dv = r 3 dr, then du = 1 r dr and v = dv = 1 4 r 4. r 3 lnr = 1 4 r 4 1 r dr r 3 dr = 1 4 r 4 lnr = 1 4 r 4 lnr 1 4 = 1 4 r 4 lnr 1 16 r 4 +C Then 3 1 r 3 lnr = [ 1 4 34 ln3 1 16 34 ] [ 1 4 14 ln1 1 16 14 ] = 81 4 ln3 5 9 3 1 r 3 lnr = 9[ 81 4 729 ln3 5] = 4 ln3 45 7
Integration by Parts: Let u = cos 1 x, dv = dx, then du = 1 1 x 2 dx and v = dv = x. cos 1 xdx = x = cos 1 (x)x + = cos 1 (x)x + 1 dx 1 x 2 x 1 x 2 dx To find x 1 x 2 dx, use u-substitution. Let u = 1 x2,then du = 2xdx or xdx = 1 2 du x dx = 1 1 x 2 2 1 du = 1 u 2 2 u = 1 x 2 Plug back to the first part 7 cos 1 xdx = 7[ 1 π 2 3 3 4 + 1] = 7π 6 7 3 4 + 7 cos 1 xdx = cos 1 (x)x 1 x 2 = 1 π 3 2 3 4 + 1 8
Integration by Parts: Let u = r 2, dv = r 16+r 2 dr, then du = 2rdr and v = dv = 16 + r 2. r 3 16 + r 2 dr = = r 2 16 + r 2 16 + r 2 2rdr We can use u-substitution to find 16 + r 2 2rdr Let u = 16 + r 2, then du = 2rdr, 16 + r 2 2rdr = udu = 2 3 u 3 2 = 2 3 (16 + r 2 ) 3 2. Plug back to the previous step: r 3 16 + r 2 dr = r 2 16 + r 2 16 + r 2 2rdr = r 2 16 + r 2 2 3 (16 + r 2 ) 3 2 1 0 r 3 16 + r 2 dr = 12 16 + 1 2 2 3 (16 + 12 ) 3 2 [0 2 16 + 0 2 2 3 (16 + 02 ) 3 2 ] = 17 2 3 (17) 3 2 + 2 3 (16) 3 2 = 128 3 31 3 17 9
Integration by Parts: We do u-substitution first. Here I am going to use y instead of u. Let y = x, then d y = 1 2 x dx or dx = 2 xd y = 2yd y 7 cos xdx = 7 (cos y)2yd y = 14 y cos yd y Now use integration by parts: Let u = y, dv = cos(x), then du = d y and v = dv = cos(y)d y = sin(y). 7 cos xdx = 7 (cos y)2yd y = 14 y cos yd y = 14 udv = 14[uv ] = 14[y sin(y) sin(y)d y] = 14[y sin y + cos y] = 14[ x sin x + cos x] +C 10
Integration by Parts: We do u-substitution first. Here I am going to use y instead of u. Let y = x + 3, then d y = dx and x = y 3 x ln(3 + x)dx = (y 3)ln yd y Now use integration by parts: Let u = ln y, dv = (y 3), then du = 1 y d y and v = dv = (y 3)d y = 1 2 y 2 3y. x ln(3 + x)dx = (y 3)ln yd y = udv = [uv ] = ( 1 2 y 2 3y)ln y ( 1 2 y 2 3y) 1 y d y = ( 1 2 y 2 3y)ln y ( 1 2 y 3)d y = ( 1 2 y 2 3y)ln y 1 4 y 2 + 3y +C = [ 1 2 (x + 3)2 3(x + 3)]ln(x + 3) 1 4 (x + 3)2 + 3(x + 3) +C 11
Integration by Parts: let u = x, dv = f (x)dx, then du = dx and v = f (x)dx = f (x). x f (x)dx = udv = uv = x f (x) f (x)dx = x f (x) f (x) 4 1 x f (x)dx = 4f (4) f (4) [1f (1) f (1)] = 4 3 9 [1 3 1] = 1 12