SUBSTITUTION METHOD IN INTEGRATION

In this page substitution method in integration we are going see where we need to use this method in integration.

In this method we need to change the function which is defined one variable to another variable. If we have limits then we have to change that too.

Now let us see some example problems to understand this topic.

Problem 1 :

Integrate cos x/(1+sin x)

Solution :

=  ∫[cos x/(1 + sin x)] dx

Let U = 1+sin x

Differentiate with respect to x on both sides

du  =  cos x dx

∫ [cos x/(1+sin x)] dx  =  ∫ [cos x dx/(1+sin x)]

=  ∫ du/u

=  ∫ (1/u) du

=  log u + C

=  log (1+sin x) + C

Problem 2 :

Integrate (6x + 5)/√(3x²+5x+6)

Solution :

=  ∫ [(6x + 5)/√(3x²+5x+6)] dx

Let U = 3x² + 5x + 6

Differentiate with respect to x on both sides

du = (6x+5) dx

=  ∫ (6x+5) dx/√(3x²+5x+6)

=  ∫du/√u

=  ∫ u^(-1/2) du

=  u^[(-1/2) + 1]/[(-1/2) + 1] + C

=  u^[(-1 + 2)/2]/[(-1 + 2)/2] + C

=  u^(1/2)/(1/2) + C

=  2 √u + C

=  2 √(3x²+5x+6) + C

Problem 3 :

Integrate cosec x

Solution :

=  ∫ cosec x dx

To solve this problem we have to multiply and divide by (cosec x - cot x)

=  ∫ cosec x (cosec x - cot x)/(cosec x - cot x) dx

Let U = (cosec x - cot x)

Differentiate with respect to x on both sides

du = - cosec x cot x - (- cosec² x)dx

du = - cosec x cot x + cosec² x)dx

du =   cosec²x- cosec x cot x dx

=  ∫ (cosec²x - cosec x cot x) dx /(cosec x - cot x)

=  ∫ du/u

=  ∫ (1/u) du

=  log u + C

=  log (cosec x - cot x) + C

Problem 4 :

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 5 :

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 6 :

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

Problem 7 :

∫ 3t² (t³ + 4)⁵ dt

Solution :

∫3t² (t³ + 4)⁵ dt

Let u = t³ + 4

du = (3t² + 0) dt

du = 3t² dt

∫ 3t² (t³ + 4)⁵ dt = ∫u⁵ du

= u⁶/6 + C

= (t³ + 4)⁶ / 6 + C

Problem 8 :

∫ √(4x - 5) dx

Solution :

∫ √(4x - 5) dx

Let t = 4x - 5

dt = 4 dx

dx = (1/4) dt

∫ √(4x - 5) dx = ∫ √t (1/4) dt

= (1/4)∫ √t dt

= (1/4) ∫t^1/2 dt

= (1/4) t^{(1/2) + 1} / [(1/2) + 1] + C

= (1/4) t^(3/2) / (3/2) + C

= (1/4) t^(3/2) x (2/3) + C

= (1/6) t^(3/2) + C

= (1/6) (4x - 5)^(3/2) + C

Problem 9 :

∫ t² (t³ + 4)^-1/2 dt

Solution :

∫ t² (t³ + 4)^-1/2 dt

Let u = t³ + 4

du = 3t² dt

t² dt = (1/3) du

∫ t² (t³ + 4)^-1/2 dx = ∫ u^-1/2 (1/3) du

= (1/3)∫ u^-1/2 du

Applying the value of u.

= (1/3) u ^1/2/(1/2) + C

Problem 10 :

∫ sin¹⁰ x cos x dx

Solution :

∫ sin¹⁰ x cos x dx

Let sin x = t

cos x dx = dt

∫ sin¹⁰ x cos x dx = ∫ t¹⁰ dt

= t¹¹/11 + C

Applying the value of t.

= sin¹¹ x / 11 + C

Problem 11 :

∫ (sin x/cos⁵ x) dx

Solution :

∫ (sin x/cos⁵ x) dx

Let cos x = t

-sin x dx = dt

sin x dx = -dt

∫ (sin x/cos⁵ x) dx = ∫ (-dt/t⁵)

= -∫ (1/t⁵) dt

= - ∫t^-5 dt

= - (t^{-5 + 1}) / (-5 + 1) + C

= - (t^-4) / (-4) + C

= - (t^-4) / (-4) + C

= - 1/4t⁴ + C

Applying the value of t, we get

= (- 1/4)cos⁴x + C

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