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Chapter 6 - SECTION 6.1(PAGE 321 R A ADAMS CALCULUS CHAPTER...

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SECTION 6.1 (PAGE 321) R. A. ADAMS: CALCULUS CHAPTER 6. TECHNIQUES OF INTE- GRATION Section 6.1 Integration by Parts (page 321) 1. x cos x dx U = x dU = dx dV = cos x dx V = sin x = x sin x sin x dx = x sin x + cos x + C . 2. ( x + 3 ) e 2 x dx U = x + 3 dU = dx dV = e 2 x dx V = 1 2 e 2 x = 1 2 ( x + 3 ) e 2 x 1 2 e 2 x dx = 1 2 ( x + 3 ) e 2 x 1 4 e 2 x + C . 3. x 2 cos π x dx U = x 2 dU = 2 x dx dV = cos π x dx V = sin π x π = x 2 sin π x π 2 π x sin π x dx U = x dU = dx dV = sin π x dx V = − cos π x π = x 2 sin π x π 2 π x cos π x π + 1 π cos π x dx = 1 π x 2 sin π x + 2 π 2 x cos π x 2 π 3 sin π x + C . 4. ( x 2 2 x ) e kx dx U = x 2 2 x dU = ( 2 x 2 ) dx dV = e kx V = 1 k e kx = 1 k ( x 2 2 x ) e kx 1 k ( 2 x 2 ) e kx dx U = x 1 dU = dx dV = e kx dx V = 1 k e kx = 1 k ( x 2 2 x ) e kx 2 k 1 k ( x 1 ) e kx 1 k e kx dx = 1 k ( x 2 2 x ) e kx 2 k 2 ( x 1 ) e kx + 2 k 3 e kx + C . 5. x 3 ln x dx U = ln x dU = dx x dV = x 3 dx V = x 4 4 = 1 4 x 4 ln x 1 4 x 3 dx = 1 4 x 4 ln x 1 16 x 4 + C . 6. x ( ln x ) 3 dx = I 3 where I n = x ( ln x ) n dx U = ( ln x ) n dU = n x ( ln x ) n 1 dx dV = x dx V = 1 2 x 2 = 1 2 x 2 ( ln x ) n n 2 x ( ln x ) n 1 dx = 1 2 x 2 ( ln x ) n n 2 I n 1 I 3 = 1 2 x 2 ( ln x ) 3 3 2 I 2 = 1 2 x 2 ( ln x ) 3 3 2 1 2 x 2 ( ln x ) 2 2 2 I 1 = 1 2 x 2 ( ln x ) 3 3 4 x 2 ( ln x ) 2 + 3 2 1 2 x 2 ( ln x ) 1 2 I 0 = 1 2 x 2 ( ln x ) 3 3 4 x 2 ( ln x ) 2 + 3 4 x 2 ( ln x ) 3 4 x dx = x 2 2 ( ln x ) 3 3 2 ( ln x ) 2 + 3 2 ( ln x ) 3 4 + C . 7. tan 1 x dx U = tan 1 x dU = dx 1 + x 2 dV = dx V = x = x tan 1 x x dx 1 + x 2 = x tan 1 x 1 2 ln ( 1 + x 2 ) + C . 8. x 2 tan 1 x dx U = tan 1 x dU = dx 1 + x 2 dV = x 2 dx V = x 3 3 = x 3 3 tan 1 x 1 3 x 3 1 + x 2 dx = x 3 3 tan 1 x 1 3 x x 1 + x 2 dx = x 3 3 tan 1 x x 2 6 + 1 6 ln ( 1 + x 2 ) + C . 212

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INSTRUCTOR’S SOLUTIONS MANUAL SECTION 6.1 (PAGE 321) 9. x sin 1 x dx U = sin 1 x dU = dx 1 x 2 dV = x dx V = x 2 2 = 1 2 x 2 sin 1 x 1 2 x 2 dx 1 x 2 Let x = sin θ dx = cos θ d θ = 1 2 x 2 sin 1 x 1 2 sin 2 θ d θ = 1 2 x 2 sin 1 x 1 4 sin θ cos θ) + C = 1 2 x 2 1 4 sin 1 x + 1 4 x 1 x 2 + C . 10. x 5 e x 2 dx = I 2 where I n = x ( 2 n + 1 ) e x 2 dx U = x 2 n dU = 2 nx ( 2 n 1 ) dx dV = xe x 2 dx V = − 1 2 e x 2 = − 1 2 x 2 n e x 2 + n x ( 2 n 1 ) e x 2 dx = − 1 2 x 2 n e x 2 + nI n 1 I 2 = − 1 2 x 4 e x 2 + 2 1 2 x 2 e x 2 + xe x 2 dx = − 1 2 e x 2 ( x 4 + 2 x 2 + 2 ) + C . 11. I n = π/ 4 0 sec n x dx U = sec n 2 x dU = ( n 2 ) sec n 2 x tan x dx dV = sec 2 x dx V = tan x = tan x sec n 2 x π/ 4 0 ( n 2 ) π/ 4 0 sec n 2 x tan 2 x dx = ( 2 ) n 2 ( n 2 )( I n I n 2 ). ( n 1 ) I n = ( 2 ) n 2 + ( n 2 ) I n 2 . Therefore I n = ( 2 ) n 2 n 1 + n 2 n 1 I n 2 , ( n 2 ). For n = 5 we have π/ 4 0 sec 5 x dx = I 5 = 2 2 4 + 3 4 I 3 = 2 2 + 3 4 2 2 + 1 2 I 1 = 7 2 8 + 3 8 ln | sec x + tan x | π/ 4 0 = 7 2 8 + 3 8 ln ( 1 + 2 ).
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Chapter 6 - SECTION 6.1(PAGE 321 R A ADAMS CALCULUS CHAPTER...

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