INTEGRATION USING SUBSTITUION METHOD PRACTICE PROBLEMS

Problem 1 :

Integrate x/√(x²+3)

Solution :

Let t  =  x²+3

differentiate with respect to x

dt  =  2x dx

x dx  =  dt/2

=  ∫ x/√(x²+3) dx

=  ∫ (dt/2)√t

=  (1/2) ∫√t dt

=  (1/2) ∫t^(1/2) dt

=  (1/2) (t^3/2 / (3/2) + C

=  (1/2) (2/3) t^3/2 + C

=  (1/3) (x²+3)^3/2 + C

Problem 2 :

Integrate (2 x + 3) √(x²+3x-5)

Solution :

Let t  =  x²+3x-5

differentiate with respect to x

dt  =  (2x+3) dx

=  ∫(2x+3) √(x²+3x-5) dx

=  ∫(dt/√t)

=  ∫t^(-1/2) dt

=  t^(1/2)/(1/2) + C

=  2√(x²+3x-5) + C

Problem 3 :

Integrate tan x

Solution :

=  ∫tan x dx

=  ∫(sin x/cos x) dx

t = cos x

differentiating with respect to x, we get

dt  =  -sin x dx

sin x dx  =  - dt

=  ∫(-dt/t)

=  -∫(1/t) dt

=  -log t + C

=  -log (cos x) + C

Problem 4 :

Integrate sec x

Solution :

=  ∫sec x dx

=  ∫ [sec x (sec x+tan x)/(sec x+tan x)] dx

=  ∫[(sec² x + sec x tan x)/(sec x + tan x)] dx

t  =  sec x + tan x

dt  =  sec x tan x + sec ² x

=  ∫dt/t

=  ∫(1/t) dt

=  log t + C

=  log (sec x + tan x) + C

Problem 5 :

Integrate x⁵ (1 + x⁶)⁷

Solution :

Let t = 1 + x⁶

differentiate with respect to x

dt  =  6 x⁵ dx

dt/6  =  x⁵ dx

 x⁵ dx  =  dt/6

=  ∫ x⁵ (1 + x⁶)⁷ dx

=  ∫ t⁷ (dt/6)

=  (1/6) t⁷ dt

=  (1/6) [t⁽⁷⁺¹⁾/(7+1)] + C

=  (1/6) (t⁸/8) + C

=  (1/48) t⁸ + C

=  t⁸/48 + C    

= (1 + x⁶)⁸/48 + C

Problem 6 :

Integrate (2Lx + m)/(Lx² + mx + n)

Solution :

Let t  =  (Lx²+mx+n)

differentiate with respect to x

dt  =  (2Lx + m) dx

=  ∫ (dt/t)

=  log t + C

=  log (Lx² + mx + n) + C

Problem 7 :

Integrate (4ax + 2b)/(ax² + bx + c)¹⁰

Solution :

t  =  ax²+bx+c

differentiate with respect to x

dt  =  2ax+b

= ∫(4ax+2b)/(ax²+bx+c)¹⁰ dx

now we are going to take 2 from the numerator

=  ∫ 2 (2ax+2b)/(ax²+bx+c)¹⁰ dx

= ∫2 (dt/t¹⁰)

=  ∫2 t^-10 dt

=  2t^(-10+1)/(-10+1) + C

=  2t^-9/(-9) + C

=  (-2/9) (ax² + bx + c)^(-9) + C

=  [-2/9(ax²+bx+c)⁹] + C

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