INTEGRATION BY SUBSTITUTION  EXAMPLES WITH SOLUTIONS

The method of substitution in integration is similar to finding the derivative of function of function in differentiation. By using a suitable substitution, the variable of integration is changed to new variable of integration which will be integrated in an easy manner.

Question 1 :

Integrate the following with respect to x

α β x^{α - 1} e^{-β x^α}

Solution :

=  ∫[α β x^{α - 1} e^{-β x^α}] dx 

Let u = βx^α

du  =  αβ x^{(α - 1)} dx

∫[α β x^{α - 1} e^{-β x^α}] dx  =   ∫e^-u du

  =  - e^-u + c

  =  - e^(^-βx^α) + c

Question 2 :

Integrate the following with respect to x

tan x √sec x

Solution :

=  ∫[tan x √sec x] dx 

Let u = √sec x

u²  =  sec x

2u du  =  sec x tan x dx

2u du/sec x  =  tan x dx

∫[tan x √sec x] dx  =  ∫(2u du/sec x) u

  =  ∫(2u/u²) u du

  =  2 ∫du

=  2 u + c

=  2  √sec x + c

Question 3 :

Integrate the following with respect to x

x (1 - x)¹⁷

Solution :

=  ∫[x (1 - x)¹⁷] dx 

Let u  =  (1 - x)

x  =  1 - u

dx  =  -du

∫[x (1 - x)¹⁷] dx  =  ∫(1-u) u¹⁷ (-du) 

 =  - ∫(u¹⁷ - u¹⁸) du

 =  - [u¹⁸/18 - u¹⁹/19]

 =  ((1 - x)¹⁹/19) - ((1 - x)¹⁸/18) + c

Question 4 :

Integrate the following with respect to x

sin⁵x cos³x

Solution :

=  ∫[sin⁵x cos³x] dx 

=  ∫[sin⁵x cos²x cos x] dx 

=  ∫[sin⁵x (1-sin²x) cos x] dx 

Let u = sin x

du  =  cos x dx

∫[sin⁵x (1-sin²x) cos x] dx =  ∫u⁵ (1-u²) du 

  =  ∫(u⁵ - u⁷) du 

  =  u⁶/6 - u⁸/8 + c

  =  (sin⁶x/6) - (sin⁸x/8) + c

Question 5 :

Integrate the following with respect to x

cos x/cos (x-a)

Solution :

=  ∫[cos x/cos (x-a)] dx 

Let u = x - a

du  =  dx 

x  =  u + a

cos x   =  cos (u + a)

cos x/cos (x-a)  =  cos (u + a) / cos u

  =  (cos u cos a + sin u sin a) / cos u

  =  cos a + tan u sin a

  =  ∫(cos a + tan u sin a) du

  =  cos a u + log sec u sin a  + c

  =  cos a (x - a) + log sec (x - a) sin a  + c

  =  (x - a) cos a + sin a log sec (x - a) + c

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