Higher Engineering Mathematics

402 INTEGRAL CALCULUS Sincex=asinθ, then sinθ= x a andθ=sin−^1 x a . Hence ∫ 1 √ (a^2 −x^2 ) dx=sin−^1 x a +c Problem 14. Evalua ...

INTEGRATION USING TRIGONOMETRIC AND HYPERBOLIC SUBSTITUTIONS 403 H Determine ∫ √ (16− 9 t^2 )dt [ 8 3 sin−^1 3 t 4 + t 2 √ (16 ...

404 INTEGRAL CALCULUS Letx=asinhθ, then dx dθ =acoshθand dx=acoshθdθ Hence ∫ 1 √ (x^2 +a^2 ) dx = ∫ 1 √ (a^2 sinh^2 θ+a^2 ) (aco ...

INTEGRATION USING TRIGONOMETRIC AND HYPERBOLIC SUBSTITUTIONS 405 H = ∫ √ (a^2 cosh^2 θ)(acoshθdθ), since cosh^2 θ−sinh^2 θ= 1 = ...

406 INTEGRAL CALCULUS which provides as alternative solution to ∫ 1 √ (x^2 −a^2 ) dx Problem 25. Determine ∫ 2 x− 3 √ (x^2 −9) d ...

INTEGRATION USING TRIGONOMETRIC AND HYPERBOLIC SUBSTITUTIONS 407 H Hence ∫ 3 2 √ (x^2 −4) dx = [ 3 2 √ 5 −2(0.9624) ] −[0] = 1. ...

Integral calculus 41 Integration using partial fractions 41.1 Introduction The process of expressing a fraction in terms of simp ...

INTEGRATION USING PARTIAL FRACTIONS 409 H By dividing out and resolving into partial fractions it was shown in Problem 4, page 2 ...

410 INTEGRAL CALCULUS Thus ∫ 2 x+ 3 (x−2)^2 dx≡ ∫ { 2 (x−2) + 7 (x−2)^2 } dx =2ln(x−2)− 7 (x−2) +c ⎡ ⎣ ∫ 7 (x−2)^2 dxis determin ...

INTEGRATION USING PARTIAL FRACTIONS 411 H It was shown in Problem 9, page 23: 3 + 6 x+ 4 x^2 − 2 x^3 x^2 (x^2 +3) ≡ 2 x + 1 x^2 ...

412 INTEGRAL CALCULUS = 1 2 a [−ln (a−x)+ln (a+x)]+c = 1 2 a ln ( a+x a−x ) +c Problem 12. Evaluate ∫ 2 0 5 (9−x^2 ) dx, correct ...

H Integral calculus 42 The t=tan θ 2 substitution 42.1 Introduction Integrals of the form ∫ 1 acosθ+bsinθ+c dθ, wherea,bandcare ...

414 INTEGRAL CALCULUS = ∫^1 2 t 1 +t^2 ( 2 dt 1 +t^2 ) = ∫ 1 t dt=lnt+c Hence ∫ dθ sinθ =ln ( tan θ 2 ) +c Problem 2. Determine: ...

THEt=tanθ 2 SUBSTITUTION 415 H from 12 of Table 40.1, page 398. Hence ∫ dθ 5 +4 cosθ = 2 3 tan−^1 ( 1 3 tan θ 2 ) +c Now try the ...

416 INTEGRAL CALCULUS from 12, Table 40.1, page 398. Hence ∫ dx 7 −3 sinx+6 cosx =tan−^1 ⎛ ⎜ ⎝ tan x 2 − 3 2 ⎞ ⎟ ⎠+c Problem 7. ...

Assign-11-H8152.tex 23/6/2006 15: 12 Page 417 H Integral calculus Assignment 11 This assignment covers the material contained in ...

Integral calculus 43 Integration by parts 43.1 Introduction From the product rule of differentiation: d dx (uv)=v du dx +u dv dx ...

INTEGRATION BY PARTS 419 H = 3 2 te^2 t− 3 2 ∫ e^2 tdt = 3 2 te^2 t− 3 2 ( e^2 t 2 ) +c Hence ∫ 3 te^2 tdt=^32 e^2 t ( t−^12 ) + ...

420 INTEGRAL CALCULUS From Problem 1, ∫ xcosxdx=xsinx+cosx Hence ∫ x^2 sinxdx =−x^2 cosx+ 2 {xsinx+cosx}+c =−x^2 cosx+ 2 xsinx+2 ...

INTEGRATION BY PARTS 421 H Letu=lnx, from which du= dx x and let dv= √ xdx=x 1 (^2) dx, from which, v= ∫ x 1 (^2) dx= 2 3 x 3 2 ...