Higher Engineering Mathematics, Sixth Edition
402 Higher Engineering Mathematics ∫ 1 3 cos5xsin2xdx = 1 3 ∫ 1 2 [sin( 5 x+ 2 x)−sin( 5 x− 2 x)]dx, from 7 of Table 40.1 = 1 6 ...
Integration using trigonometric and hyperbolic substitutions 403 = ∫ acosθdθ √ (a^2 cos^2 θ) ,sincesin^2 θ+cos^2 θ= 1 = ∫ acosθd ...
404 Higher Engineering Mathematics Determine ∫ √ ( 16 − 9 t^2 )dt. [ 8 3 sin−^1 3 t 4 + t 2 √ ( 16 − 9 t^2 )+c ] Evaluate ∫ ...
Integration using trigonometric and hyperbolic substitutions 405 Hence ∫ 1 √ (x^2 +a^2 ) dx = ∫ 1 √ (a^2 sinh^2 θ+a^2 ) (acoshθd ...
406 Higher Engineering Mathematics = a^2 2 ( θ+ sinh2θ 2 ) +c = a^2 2 [θ+sinhθcoshθ]+c, since sinh2θ=2sinhθcoshθ Sincex=asinhθ,t ...
Integration using trigonometric and hyperbolic substitutions 407 ∫ 2 x− 3 √ (x^2 − 9 ) dx= ∫ 2 x √ (x^2 − 9 ) dx − ∫ 3 √ (x^2 − ...
408 Higher Engineering Mathematics Find ∫ 3 √ ( 4 x^2 − 9 ) dx. [ 3 2 cosh−^1 2 x 3 +c ] Find ∫ √ (θ^2 − 9 )dθ. [ θ 2 √ (θ^2 ...
Chapter 41 Integration using partial fractions 41.1 Introduction The process of expressing a fraction in terms of simpler fracti ...
410 Higher Engineering Mathematics By dividing out (since the numerator and denomina- tor are of the same degree) and resolving ...
Integration usingpartial fractions 411 7. ∫ 6 4 x^2 −x− 14 x^2 − 2 x− 3 dx [0.8122] Determine the value of k, given that: ∫ 1 ...
412 Higher Engineering Mathematics Now try the following exercise Exercise 163 Further problems on integration using partial fra ...
Integration usingpartial fractions 413 = 1 2 a [ln(x−a)−ln(x+a)]+c = 1 2 a ln ( x−a x+a ) +c Problem 10. Evaluate ∫ 4 3 3 (x^2 − ...
Chapter 42 The t = tan θ 2 substitution 42.1 Introduction Integrals of the form ∫ 1 acosθ+bsinθ+c dθ,where a,bandcare constants, ...
Thet=tanθ 2 substitution 415 42.2 Worked problemson the t=tan θ 2 substitution Problem 1. Determine ∫ dθ sinθ Ift=tan θ 2 then s ...
416 Higher Engineering Mathematics If t=tan θ 2 then cosθ= 1 −t^2 1 +t^2 and dx= 2dt 1 +t^2 from equations (2) and (3). Thus ∫ d ...
Thet=tanθ 2 substitution 417 = ∫ 2dt 1 +t^2 7 ( 1 +t^2 )− 3 ( 2 t)+ 6 ( 1 −t^2 ) 1 +t^2 = ∫ 2dt 7 + 7 t^2 − 6 t+ 6 − 6 t^2 = ∫ 2 ...
418 Higher Engineering Mathematics 4. ∫ dθ 3 −4sinθ ⎡ ⎢ ⎣ 1 √ 7 ln ⎧ ⎪⎨ ⎪⎩ 3tan θ 2 − 4 − √ 7 3tan θ 2 − 4 + √ 7 ⎫ ⎪⎬ ⎪⎭ +c ⎤ ⎥ ...
Revision Test 12 This Revision Test covers the material contained in Chapters 40 to 42.The marks for each question are shown in ...
Chapter 43 Integration by parts 43.1 Introduction From the product rule of differentiation: d dx (uv)=v du dx +u dv dx , whereua ...
Integration by parts 421 Problem 2. Find ∫ 3 te^2 tdt. Letu= 3 t, from which, du dt =3, i.e. du=3dtand let dv=e^2 tdt, from whic ...
«
17
18
19
20
21
22
23
24
25
26
»
Free download pdf