INTEGRATION BY PARTS TRIGONOMETRIC SUBSTITUTION

Integration by parts method is generally used to find the integral when the integrand is a product of two different types of functions or a single logarithmic function or a single inverse trigonometric function or a function which is not integrable directly.

Formula :

∫udv  =  uv - ∫vdu

By differentiating 'u' we get 'du', by integrating dv we get v.

Practice Questions

Question 1 :

Integrate the following with respect to x:

 x sin^-1x /√(1-x²)

Solution :

  =  ∫ x sin^-1x /√(1-x²) dx

t = sin^-1x  ===>  x  =  sin t

dt  =  1/√(1-x²) dx

  =  ∫ x t dt

  =  ∫ t sin t dt

u = t            ∫dv  =  ∫ sin t dt

du  = dt          v  =  - cos t

∫ udv  =  uv - ∫vdu

  =  t(-cos t) - ∫ (-cos t) dt

  =  - t cos t + ∫ cos t dt

  =  - t cos t + sin t + c

 =  - t √(1 - sin²t) + sin t + c

 =  - sin^-1x √(1 - x²) + x + c

Question 2 :

Integrate the following with respect to x:

 x⁵ e^x² 

Solution :

  =  ∫ x⁵ e^x²  dx

  =  ∫ (x²)² x e^x² dx

Let x²  =  t

2x dx  =  dt

x dx  =  dt/2

∫ (x²)² x e^x² dx  =  ∫ t² e^t (dt/2)

  =  (1/2) ∫ t² e^t dt

u  =  t²     ∫dv  =  ∫e^t dt

u'  =  2t     v  =  e^t 

u''  =  2      v₁  =  e^t

∫ udv  =  uv - u'v₁ + u''v₂ - u'''v₃ + ...................

  =  (1/2) [t²(e^t) - 2t(e^t) + 2 (e^t)]

  =  (1/2) [e^t (t² - 2t + 2)] + c

By applying the value of "t", we get

  =  (1/2) [e^x² ((x²)² - 2(x²) + 2)] + c

  =  (1/2)e^x² (x⁴- 2x² + 2) + c

Question 3 :

Integrate the following with respect to x:

 tan^-1 (8x /1 - 16x²)

Solution :

  =  ∫  tan^-1 (8x /1 - 16x²) dx 

  =  ∫ tan^-1 (2(4x) /1 - (4x)²) dx

Let 4x  =  tan θ

4 dx  =  sec²θ dθ

  =  ∫ tan^-1 (2 tan θ /1 - tan²θ) dx

=  ∫ tan^-1 (tan 2θ) dx

=  ∫ 2θ (sec²θ/4) dθ

=  (1/2) ∫ θ sec²θ dθ

u = θ            dv  =  sec²θ

du =  dθ          v  = tan θ

∫ udv  =  uv - ∫vdu

  =  θtan θ - ∫tan θ dθ

  =  θ tan θ - log sec θ  + c

  =   4x tan^-1 (4x) - log √(1+(4x)²) + c

  =   4x tan^-1 (4x) - log √(1+16x²) + c

Question 4 :

Integrate the following with respect to x:

 sin^-1 (2x /1 + x²)

Solution :

  =  ∫sin^-1 (2x /1 + x²) dx

Let x  =  tan θ

dx  =  sec²θ dθ

  =  ∫ sin^-1 (2 tan θ /1 + (tan θ)²) dx

  =  ∫ sin^-1 (2 tan θ /(1 + tan²θ)) dx 

  =  ∫ sin^-1 (sin 2θ) dx

  =  ∫ 2θ dx

  =  ∫ 2θ sec²θ dθ

u = θ             dv  =  sec²θ dθ

du  =  1           v  =  tan θ

∫ udv  =  uv - ∫vdu

  =  2 [θtan θ - ∫tan θ dθ]

  =  2 [θ tan θ - log sec θ + c]

  =  2 [x tan^-1 x - log  √(1+tan²θ) + c

  =  2 [x tan^-1 x - log  √(1+x²) + c

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