Higher Engineering Mathematics

422 INTEGRAL CALCULUS i.e. ( 1 + a^2 b^2 )∫ eaxcosbxdx = 1 b eaxsinbx+ a b^2 eaxcosbx i.e. ( b^2 +a^2 b^2 )∫ eaxcosbxdx = eax b^ ...

INTEGRATION BY PARTS 423 H In determining a Fourier series to repre- sentf(x)=xin the range−πtoπ, Fourier coefficients are give ...

Integral calculus 44 Reduction formulae 44.1 Introduction When using integration by parts in Chapter 43, an integral such as ∫ x ...

REDUCTION FORMULAE 425 H and I 0 = ∫ x^0 exdx= ∫ exdx=ex Thus ∫ x^3 exdx=x^3 ex−3[x^2 ex− 2 I 1 ] =x^3 ex−3[x^2 ex−2(xex−I 0 )] ...

426 INTEGRAL CALCULUS From equation (2), ∫ t^3 costdt=I 3 =t^3 sint+ 3 t^2 cost−3(2)I 1 and I 1 =t^1 sint+ 1 t^0 cost−1(0)In− 2 ...

REDUCTION FORMULAE 427 H Problem 6. Use a reduction formula to deter- mine ∫ x^3 sinxdx. Using equation (3), ∫ x^3 sinxdx=I 3 =− ...

428 INTEGRAL CALCULUS du dx =(n−1) sinn−^2 xcosx and du=(n−1) sinn−^2 xcosxdx and let dv=sinxdx, from which, v= ∫ sinxdx=−cosx. ...

REDUCTION FORMULAE 429 H =4[(− 0. 054178 − 0. 1020196 − 0 .2881612)−(− 0 .533333)] =4(0.0889745)= 0. 356 Problem 10. Determine a ...

430 INTEGRAL CALCULUS Hence ∫ cos^4 xdx = 1 4 cos^3 xsinx+ 3 4 ( 1 2 cosxsinx+ 1 2 x ) = 1 4 cos^3 xsinx+ 3 8 cosxsinx+ 3 8 x+c ...

REDUCTION FORMULAE 431 H = ∫ tann−^2 xsec^2 xdx−In− 2 i.e.In= tann−^1 x n− 1 −In− 2 Whenn=7, I 7 = ∫ tan^7 xdx= tan^6 x 6 −I 5 I ...

432 INTEGRAL CALCULUS =x(lnx)n−n ∫ (lnx)n−^1 dx i.e.In=x(lnx)n−nIn− 1 Whenn=3, ∫ (lnx)^3 dx=I 3 =x(lnx)^3 − 3 I 2 I 2 =x(lnx)^2 ...

H Integral calculus 45 Numerical integration 45.1 Introduction Even with advanced methods of integration there are many mathemat ...

434 INTEGRAL CALCULUS (a) ∫ 3 1 2 √ x dx= ∫ 3 1 2 x− 1 (^2) dx ⎡ ⎢ ⎣ 2 x (− 1 2 ) 1 − 1 2 1 ⎤ ⎥ ⎦ 3 1 [ 4 x 1 2 ] 3 1 = 4 [√ x ...

NUMERICAL INTEGRATION 435 H Corresponding values of 1 1 +sinx are shown in the table below. x 1 1 +sinx 0 1.0000 π 12 (or 15◦) 0 ...

436 INTEGRAL CALCULUS Problem 4. Use the mid-ordinate rule with (a) 4 intervals, (b) 8 intervals, to evaluate∫ 3 1 2 √ x dx, cor ...

NUMERICAL INTEGRATION 437 H Now try the following exercise. Exercise 175 Further problems on the mid-ordinate rule In Problems 1 ...

438 INTEGRAL CALCULUS y y 1 y 2 y 3 y 4 y 2 n+ 1 a ddd y = f(x) Oxb Figure 45.4 ≈ 1 3 d[(y 1 +y 2 n+ 1 )+4(y 2 +y 4 +···+y 2 n) ...

NUMERICAL INTEGRATION 439 H With 6 intervals, each will have a width of π 3 − 0 6 i.e. π 18 rad (or 10◦), and the ordinates will ...

440 INTEGRAL CALCULUS 6. ∫ 4 1 4 x^3 dx (Use 6 intervals) [ (a) 1.875 (b) 2. 107 (c) 1.765 (d) 1. 916 ] 7. ∫ 6 2 1 √ (2x−1) dx ( ...

Assign-12-H8152.tex 23/6/2006 15: 12 Page 441 H Integral calculus Assignment 12 This assignment covers the material contained in ...