1

Integration

Formula Revision

2

Revision from Higher

1. Integrate (3x-7)4/3.dx

= (3x-7)7/3 +c 7/3 x3= (3x-7)7/3 +c 7

2. Integrate sin(3x+4).dx

= -cos(3x+4) +c3

3. Integrate 1 .dx (4x+3)

= ln|4x+3| +c 4

3

Integration by Substitution

1. Find ∫x4.(1+2x5)3.dx where u=1+2x5

u=1+2x5

du=10x4

dxdu=10x4dx du =dx10x4

∫x4.(1+2x5)3.dx= ∫x4.u3.dx= ∫x4.u3. du

10x4

= ∫x4.u3. 1 du 10x4

= ∫1/10u3dx= 1/10u4 +c 4= 1/40u4 +c= 1/40(1+2x5)4 +c

4

Integration by Substitution

2. Find ∫sin2xcosc.dx where u=sinx

u=sinxdu=cosxdxdu=cosxdx

∫sin2xcosx.dx= ∫u2cosxdx= ∫u2du= u3 +c 3= sin3x +c 3

5

Integration by Parts

1. Find ∫xcosx.dx

∫fg‛ = fg- ∫gf‛So,f=xg‛=cosxf‛=1g=sinx

∫xcosx.dx = xsinx - ∫sinx =xsinx-(-cosx) + c =xsinx + cosx + c

2. Find ∫x2 e3x.dx

∫fg‛ = fg- ∫gf‛So,f = x2

f ‛= 2g‛ = e3x

g = 1/3e3x

=x2.1/3e3x - ∫2x.1/3e3x

∫2x. 1/3 e3x

So,f = 2xf ‛= 2g‛ = 1/3 e3x

g = 1/9 e3x

∫fg‛ = fg- ∫f‛g =2x. 1/9 e3x - ∫2. 1/9 e3x

=2x/9. e3x - 2/27 .e3x

= x2 + 1/3. e3x - (2x/9 e3x -2/27 e3x)

= x2/3.ex - 2/9 x e3x + 2/27 e3x + c

6

Integration using Partial Fractions

1. ∫ 2x2+4x .dx (x-1)(x+1)(2x+1)

2x2+4x= A + B + C . x-1 x+1 2x+1

2x2+4x= A(x+1)(2x+1) + B(x-1)(2x+1) + C(x-1)(x+1)

Let x=1 6=6AA=1

Let x=-1 -2=2BB=-1

Let x=-1/2 2(-1/2)2+4(-1/2) = C(-1/2-1)(-1/2+1)2(1/4)-2 = C(-3/2)(1/2)-3/2 = -3/4C-6 = -3CC = -2

∫ 1 + -1 + 2 .dxx-1 x+1 2x+1=ln|x-1|-ln|x+1|+2/2ln|2x+1|+c=ln|x-1||2x-1|+c |x+1|