Integration by Substitution Formula | Sample Problems | Class 12 Math Notes Study Material Download Free PDF

The process of finding the anti-derivative of a function is the inverse process of differentiation i.e. finding integral is the inverse process of differentiation. Integration can be used to find the area or volume of a function with or without certain limits or boundaries 

It is shown as

∫g(x)dx = G(x) + C 

Explanation:

It means integral of function “g(x)” with respective “x”

G(x) represents the anti-derivative and can also state that the derivative of G(x) w.r.t x is g(x)

g(x) is the integrand on which integration is performed

dx is integrating agent

C is called integration constant

With a diagram

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Integration by Substitution Formula | Sample Problems | Class 12 Math Notes Study Material Download Free PDF

Lets us say we need to find an Area Under a Curve let the function be f(x)

The area can be found by integrating the function between the boundaries a and b

let us say ∫ f(x)dx = F(x)

the area under the curve given as F(b)-F(a) given ‘b’ as the upper limit and ‘a’ as the lower limit

Integration for some standard function

∫ a dx = ax+ C ;Where a is constant

∫ xn dx = ((xn+1)/(n+1))+C ; n≠1.

∫ sin x dx = – cos x + C.

∫ cos x dx = sin x + C.

∫ sec2 x dx = tan x + C.

Integration by Substitution

Integration of a few standard functions is given, but to find out the integrals of various functions apart from basic functions we apply different methods to bring the functions to basic functions format so that integration can be performed. One of those methods is the Integration by substitution method.

The chain rule used to perform differentiation is applied in a reverse format which is why this method is also called as reverse chain rule or u-substitution method.

In this method, the integral function is transformed into another format i.e. into the simplest form by replacing or substituting independent variables like “x” with others 

Example: ∫(3x2-5)(6x)dx

Solution:

Let us assume 3x2-5=t

performing the differentiation on both sides

6x.dx=dt

Now substitute  3x2-5 as “t”  and 6x.dx as dt in the given we get

∫t.dt 

w.k.t( i.e. We Know That)

∫ xn dx = ((xn+1)/(n+1))+C ; n ≠ 1

applying this standard formula we get (t2)/2+c

replacing t with  3x2-5

The answer is  ((3x2-5)2)/2

When to apply the Integration by Substitution method

To find integral by using this method it needs to be present in a specific format and the general form is given as

∫ f(g(x)).g'(x).dx = f(t).dt

where g(x)=t

g'(x).dx=dt

∫f(t).dt=F(t)+c=F(g(x))+c

We substitute g(x) with t and convert it into a simple or standard function format and perform the integration and finally replace t with g(x)

Sample Problems

Question 1: ∫tan x dx

Solution:

∫(sin x/cos x).dx

let us assume cos x=t

performing the differentiation on both sides

-sin x.dx=dt

=-∫dt/t

w.k.t∫1/t=ln |t| ….(w.k.t. = We Know That )

=-ln |t|

replacing t with cos x

=-ln |cos x| or ln |sec x|

Question 2: ∫cot x dx

Solution:

∫(cos x/sin x)dx

assume sin x=t

performing the differentiation on both sides

cos x.dx=dt

=∫dt/t

w.k.t∫1/t=ln|t|

=ln|t|

replacing t with sin x

=ln |sin x| or -ln |cos x|

Question 3: ∫sec x dx

Solution:

It can be rewritten as ∫(sec x (sec x+tan x))/(sec x+tan x) dx

let us assume secx+tanx=t

performing the differentiation on both sides

(sec x.tan x+sec2 x)dx=dt

sec x(tan x+sec x).dx=dt

=∫dt/t

w.k.t∫1/t=ln|t|

=ln|t|

replacing t with secx+tanx

=ln |secx+tanx|

Question 4: ∫cosec x dx

Solution:

It can be rewritten as 

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

let us assume cosec x+cot x=t

(-cosec x.cot x-cosec2 x)dx=dt

-cosec x(cot x+cosec x)dx=dt

=-∫dt/t

w.k.t∫1/t=ln |t|

=ln |t|

replacing t with cosec x +co tx

=ln |cosec x + cot x|

Question 5: ∫x.sin(8+2x2)dx

Solution:

Assume 8+2x2=t

performing the differentiation on both sides

4xdx=dt

xdx=dt/4

=1/4∫sin(t)dt

wkt ∫ sin x dx = – cos x + C.

=1/4∫sin(t)dt=1/4(-cost)+c

replacing t with 8+2(x^2)

=(-cos(8+2x2))/4+C

Question 6: ∫(2w-4)(2w2-8w+10)3.dw

Solution:

Assume 2w2-8w+10=t

performing the differentiation on both sides

(4w-8)dw=dt

(2w-4)dw=dt/2

=1/2(∫t3.dt)

∫ xn dx = ((xn+1)/(n+1))+C ; n≠1

=1/2(∫t3.dt)=(1/2).(1/4)(t4)+C

replacing t with 2w2-8w+10

=1/8((2w2-8w+10)4)+C

Question 7: ∫p/(1+5p2).dp

Solution:

Assume 1+5p2=t

performing the differentiation on both sides

10pdp=dt

pdp=dt/10

=1/10(∫dt/t)

w.r.t ∫1/p=|logp|+C

=1/10(|logt|)+C

replacing t with 1+5p2

=1/10(|log(1+5p2)|)+C

Question 8: ∫cos(8x + 8) dx 

Solution:

Assume 8x+8=t

performing the differentiation on both sides

8dx=dt

1/8(∫cost dt)

w.r.t ∫ cos x dx = sin x + C.

1/8(∫cost dt)=1/8(sint)+C

replacing t with 8x+8

1/8(sin(8x+8))+C

Question 9: ∫(3sin x).cos x.dx

Solution:

Assume sin x=t

performing the differentiation on both sides

cos x.dx=dt

=∫(3t)dt

w.r.t ∫ ax dx = (ax/ln a) + C ; a>0,  a≠1

=∫(3t)dt=(3t)/ln 3

replacing t with sinx

=(3^sinx)/ln 3

Practice Problems – Integration by Substitution Formula

Try out the given problems to master the concept of Integration by Substitution :

Q1 : ∫ [(ln (x))ⁿ/x] dx

Q2 : ∫ [(Sin⁻¹(x))³/√(1-x²)] dx

Q3 : ∫ [2x(e)^(x²)] dx

Q4 : ∫ [x/√(1+x²)] dx

Q5 : ∫ Cos(3x)Sin(3x) dx

Q6 : ∫ [2x/(1+x²)] dx

Q7 : ∫ [x√(1-x²)] dx

Q8 : ∫ 2 e^(2x) dx

Q9 : ∫ (2x + 1)/(x² + x + 1) dx

Q10 : ∫ [Sin( ln(x)) . (1/x)] dx

Hint for Q3   -> Let u = 3x and Use identity : Sin(2u) = 2 Sin(u) Cos(u) , then let v = 2u.

FAQs on Integration by Substitution

What is integration by substitution?

Integration by substitution is a method used to evaluate integrals. It involves changing the variable of integration to simplify the integral. This technique is analogous to the chain rule for differentiation.

What is the formula for integration by substitution?

The formula for integration by substitution is:

∫f(g(x))g'(x) dx = ∫f(u) du

where , u = g(x)

How do you choose the substitution u?

Choosing the substitution u typically involves identifying a function inside the integral whose derivative is also present in the integral. Often, u is set to be the inner function in a composite function.

What if the integral doesn’t match exactly after substitution?

If the integral doesn’t match exactly after substitution, you may need to manipulate the integral algebraically. For instance, you might factor out constants or split the integral into parts.

Can integration by substitution be used for definite integrals?

Yes, integration by substitution can be used for definite integrals. However, you need to change the limits of integration according to the substitution as :

What are Common Mistakes to Avoid?

  • Forgetting to change the limits of integration when dealing with definite integrals.
  • Not simplifying the integral properly after substitution.
  • Incorrectly differentiating or integrating functions.

Are there any functions that are particularly suited for substitution?

Functions involving compositions such as polynomials within trigonometric, exponential, or logarithmic functions often goes well with substitution method. Identifying an inner function whose derivative is present in the integral is key here.

Er. Neeraj K.Anand is a freelance mentor and writer who specializes in Engineering & Science subjects. Neeraj Anand received a B.Tech degree in Electronics and Communication Engineering from N.I.T Warangal & M.Tech Post Graduation from IETE, New Delhi. He has over 30 years of teaching experience and serves as the Head of Department of ANAND CLASSES. He concentrated all his energy and experiences in academics and subsequently grew up as one of the best mentors in the country for students aspiring for success in competitive examinations. In parallel, he started a Technical Publication "ANAND TECHNICAL PUBLISHERS" in 2002 and Educational Newspaper "NATIONAL EDUCATION NEWS" in 2014 at Jalandhar. Now he is a Director of leading publication "ANAND TECHNICAL PUBLISHERS", "ANAND CLASSES" and "NATIONAL EDUCATION NEWS". He has published more than hundred books in the field of Physics, Mathematics, Computers and Information Technology. Besides this he has written many books to help students prepare for IIT-JEE and AIPMT entrance exams. He is an executive member of the IEEE (Institute of Electrical & Electronics Engineers. USA) and honorary member of many Indian scientific societies such as Institution of Electronics & Telecommunication Engineers, Aeronautical Society of India, Bioinformatics Institute of India, Institution of Engineers. He has got award from American Biographical Institute Board of International Research in the year 2005.