EXAMPLES OF INTEGRATION BY PARTS

Integration by parts is one of the method basically used o find the integral when the integrand is a product of two different kind of function.                  

Formula :

∫u dv = uv-∫v du

Given Integral

∫log x dx

∫tan⁻ ¹ x dx

∫xⁿ log x dx

∫xⁿ tan⁻ ¹ x dx

∫xⁿ eax dx

∫xⁿ sin x dx

∫xⁿ cos x dx

u

log x

tan⁻ ¹ x

log x

tan⁻ ¹ x

xⁿ

xⁿ

xⁿ

dv

dx

dx

xⁿ dx

xⁿ dx

eax dx

sin x dx

cos x dx

Example 1 :

Evaluate :

∫xexdx

Solution :

u  =  x, du  =  dx

dv  =  ex dx and v  =  ex 

∫udv  =  uv-∫vdu

=  xex-exdx

=  xex-ex+C

∫ xex dx  =  ex(x-1)+C

Example 2 :

Evaluate :

∫xsinxdx

Solution :

u  =  x, du  =  dx

dv  =  sinx dx and v  =  -cosx 

∫udv  =  uv-∫vdu

=  x(-cosx)-∫(-cosx)dx

=  -xcosx+∫cosxdx

∫ x sin x dx    -xcosx + sinx + C

Example 3 :

Evaluate :

∫xlogxdx

Solution :

u  =  logx, du  =  (1/x) dx

dv  =  dx and v  =  x2/2

∫udv  =  uv-∫vdu

=  log x(x2/2) - (x2/2)(1/x)dx

=  log x(x2/2) - (1/2)xdx

=  log x(x2/2) - (1/2)(x2/2) + C

∫ x log x dx  =  (x2/2)log x - (x2/4) + C

Example 4 :

Evaluate :

∫xsec2xdx

Solution :

u  =  x, du  =  dx

dv  =  sec2x and v  =  tan x

∫udv  =  uv-∫vdu

=  x(tan x) - ∫tan x dx

∫ xsec2x dx=  x tanx - ln(secx) + C

Example 5 :

Evaluate :

∫xtan-1xdx

Solution :

u  =  x, du  =  (x2/2)dx

dv  =  tan-1 x dx and v  =  1/(1+x2)

∫udv  =  uv-∫vdu

Example 6 :

Evaluate :

∫logxdx

Solution :

u  =  logx , du  =  1/x

dv  =  dx and v  =  x

∫udv  =  uv-∫vdu

=  logx(x) - ∫xdx

∫ log x dx  =  xlogx - (x2/2) + C

Example 7 :

Evaluate :

∫sin-1xdx

Solution :

u  =  sin-1x , du  =  1/√(1-x2) dx

dv  =  dx and v  =  x

∫udv  =  uv-∫vdu

=  xsin-1x-∫x (1/√(1-x2)) dx

=  xsin-1x-∫ (x/√(1-x2)) dx

Here we may use substitution method to find integration.

Example 8 :

Evaluate :

∫xsin2xdx

Solution :

sin2x  =  (1-cos2x)/2

∫x sin2x dx  =  ∫x (1-cos2x)/2 dx

=  (1/2)[∫x(1-cos2x) dx]

=  (1/2)[∫x dx - ∫x cos2x dx]

=  (1/2)[(x2/2) - ∫x cos2x dx]  ---(1)

Integrating x cos2x :

u  =  x and du  =  dx

dv  =  cos 2x and v  =  sin2x/2

∫udv  =  uv-∫vdu

=  x(sin2x/2)-∫(sin2x/2) dx

=  (x/2)sin2x-(1/2)∫sin2x dx

∫x cos2x dx  =  (x/2)sin2x+(1/4)cos 2x

Applying the integrated answer in (1), we get

=  (1/2)(x2/2) - 1/2[(x/2)sin2x+(1/4)cos 2x] + C

=  (x2/4) - (x/4)sin2x-(1/8)cos 2x + C

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