Integration by Substitution Formula

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 finding 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) represent the anti-derivative and can also state that 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

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

∫ 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|

=-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)

wkt ∫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.k.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: ∫(3^{sin x}).cos x.dx

Solution:

Assume sin x=t

performing the differentiation on both sides

cos x.dx=dt

=∫(3^t)dt

wkt ∫ a^x dx = (a^x/ln a) + C ; a>0,  a≠1

=∫(3^t)dt=(3^t)/ln 3

replacing t with sinx

=(3^sinx)/ln 3