PRACTICE PROBLESM ON INTEGRATION BY PARTS

Evaluate the following.

Problem 1 :

∫x e^x dx

Solution :

u  =  x, du  =  dx

dv  =  e^x dx and v  =  e^x 

∫udv  =  uv-∫vdu

=  xe^x-∫e^xdx

=  xe^x-e^x+C

∫ xe^x dx  =  e^x(x-1)+C

Problem 2 :

∫x sin x dx

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

Problem 3 :

∫ x log x dx

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

Problem 4 :

∫ xsec²x dx

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

Problem 5 :

∫ x  tan^-1 x dx

Solution :

u  =  tan^-1 x, du  =  1 / (1 + x²)dx

dv = x dx and v = x²/2

∫udv  =  uv-∫vdu

= tan^-1 x (x²/2) - ∫x²/2 du

= tan^-1 x (x²/2)  - ∫1/(1+x²) (x²/2)dx

= tan^-1 x (x²/2)  - (1/2) ∫ x²/(1+x²) dx

Let us add and subtract 1 with the numerator.

= tan^-1 x (x²/2)  - (1/2) ∫ (x² + 1 - 1) / (1+x²) dx

= tan^-1 x (x²/2)  - (1/2) ∫ (x² + 1) / (1+x²) - (1 / (1+x²)) dx

= tan^-1 x (x²/2)  - (1/2) [∫ 1 dx - ∫ (1 / (1+x²)) dx]

∫ x tan^-1 x dx = tan^-1 x (x²/2)  - (1/2) [x - tan^-1 x] + C

Problem 6 :

∫ log x dx

Solution :

u  =  logx , du  =  1/x

dv  =  dx and v  =  x

∫udv  =  uv-∫vdu

=  logx(x) - ∫xdx

∫ log x dx  =  xlogx - (x²/2) + C

Problem 7 :

∫ x sin^-1x dx

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 ---(1)

Here we may use substitution method to find integration.

Let t = √(1-x²)

t² = 1 - x²

2t dt = -2x dx

x dx = -t dt

Applying in (1), we get

=  xsin^-1x-∫ (1/t) (-t dt)

=  xsin^-1x + ∫ dt

=  xsin^-1x + t + C

Applying the value of t, we get

=  xsin^-1x + √(1-x²) + C

Problem 8 :

∫x sin²x dx

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

Problem 9 :

∫x sin 3x cos 2x dx

Solution :

∫x sin 3x cos 2x dx  =  ∫x (2/2)sin 3x cos 2x dx

=  (1/2)∫x 2sin 3x cos 2x dx

=  (1/2)∫x (sin 5x+sinx)dx

=  (1/2) [∫x sin 5x dx +∫x sinx dx]

We have to use integration by parts two times.

=  (1/2) [x(-cos 5x/5) - ∫(-cos 5x/5) dx + x(-cos x) -  ∫(-cos x) dx]

=  (1/2) [x(-cos 5x/5) + (1/5)∫cos 5x - x cos x + ∫cos x dx]

=  (1/2) [(-x/5)cos 5x + (1/5)(sin 5x/5) - x cos x + sin x + C

=  (1/2) [(-x/5)cos 5x + (1/25)(sin 5x) - x cos x + sin x + C

Problem 10 :

∫x³ e^x² dx

Solution :

∫x³ e^x² dx  =  ∫x ⋅ x² e^x² dx

Let t  =  x² 

dt  =  2x dx

x dx  =  dt/2

=  ∫te^t (dt/2)

=  (1/2) ∫ te^t dt

Using the formula integrating by parts, we get

=  (1/2) ∫ te^t dt

=  (1/2) [te^t- ∫e^t dt]

=  (1/2)e^t(t-1) + C

Applying the value of t, we get

=  (1/2)e^{x^2}(x²-1) + C

Problem 11 :

∫√x ln x dx

Solution :

∫√x ln x dx

u = ln x, du = 1/x dx

dv = √x, v = x^(3/2) / (3/2)

= (2/3) x^(3/2)

= (2/3) x √x

∫uv = uv - ∫v du

= ln x (2/3) x √x -  ∫(2/3) x √x  (1/x) dx

= ln x (2/3) x^(3/2) -  (2/3)[x^(3/2)/(3/2)]

= ln x (2/3) x^(3/2) -  (2/3)(2/3) x √x + C

= (2/3) x √x ln x -  (4/9) x √x + C

Problem 12 :

∫x² cos 3x dx

Solution :

∫x² cos 3x dx

u = x², du = 2x dx

dv = cos 3x, v = sin 3x/3

∫uv = uv - ∫v du

= x²(sin 3x/3) - ∫(sin 3x/3) 2x dx

= x²(sin 3x/3) - 2/3∫ x sin 3x dx

Using integration by parts again.

u = x, du = dx

dv = sin 3x, v = - cos 3x / 3

= x²(sin 3x/3) - 2/3[x (-cos 3x/3) - ∫ (-cos 3x/3) dx 

= x²(sin 3x/3) + (2/9) (x cos 3x) - (2/9) ∫ cos 3x

= x²(sin 3x/3) + (2/9) (x cos 3x) - (2/27) sin 3x + C

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