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 = 3tdt

∫ 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) t1/2 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

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Digital SAT Math Problems and Solutions (Part - 57)

    Oct 22, 24 06:52 AM

    Digital SAT Math Problems and Solutions (Part - 57)

    Read More

  2. Solving Exponential Equations Problems and Solutions (Part - 6)

    Oct 21, 24 03:23 AM

    Solving Exponential Equations Problems and Solutions (Part - 6)

    Read More

  3. SAT Math Resources (Videos, Concepts, Worksheets and More)

    Oct 18, 24 09:12 AM

    SAT Math Resources (Videos, Concepts, Worksheets and More)

    Read More