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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 = 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 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 (θ − 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 ....
View Full Document

Page1 / 52

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online