Integration by Substitution:

Integration by substitution, also known as u-substitutionreverse chain rule or change of variables, is a method for evaluating integrals and antiderivatives. It is the counterpart to the chain rule for differentiation, and can loosely be thought of as using the chain rule "backwards".

The General Form of integration by substitution is:

Usually the method of integration by substitution is extremely useful when we make a substitution for a function whose derivative is also present in the integrand. Doing so, the function simplifies and then the basic formulas of integration can be used to integrate the function.

Example:

∫sin (x3).3x2.dx = ∫sin t . dt    Let x3= t => 3x2dx =dt

= -cos t + c

Example :

Find the integration of

Solution:

Given : I =

Example :

Integrate 6x cos (x2 – 5) with respect to x .

Solution:

I = ∫6xcos(x2 – 5).dx

Let x2 – 5 = t …..(1)

2x.dx = dt

 we have

I = 3∫cos(t).dt

= 3sin t + c …..(2)

= 3sin (x2 – 5) + C

We know that tan x = sin x / cos x.

Therefore,
∫ tan x dx = ∫ (sin x / cos x) dx.
Now, let  cos x = t, => sin x dx = – dt.

Therefore,
∫ tan x dx = – ∫ (dt / t) = – log |cos x| + C
Or, ∫ tan x dx = log |sec x| + C

We know that cot x = cos x / sin x. Therefore,
∫ cot x dx = ∫ (cos x / sin x) dx.
Now, let sin x = t, =>  cos x dx = – dt.

Therefore,
∫ cot x dx = ∫ (dt / t) = log |t| + C = log |sin x| + C

On multiplying both the numerator and denominator by (sec x + tan x), we have
∫ sec x dx = ∫ {sec x (sec x + tan x) dx} / (sec x + tan x)
Now, let (sec x + tan x) = t, => sec x (sec x + tan x) dx = dt.
Therefore, ∫ sec x dx = ∫ (dt / t) = log |t| + C = log |sec x + tan x| + C

On multiplying both the numerator and denominator by (cosec x + cot x), we have
∫ cosec x dx = ∫ {cosec x (cosec x + cot x) dx} / (cosec x + cot x)
Now, let (x + cot x) = t, =>  – cosec x (cosec x + cot x) dx = dt.
Therefore, ∫ cosec x dx = – ∫ (dt / t) = – log |t| + C = – log |cosec x + cot x| + C
= – log |(cosec2 x – cot2 x) / (cosec x – cot x)| + C
= log |cosec x – cot x| + C

Example :

Find the integral of (sin3 x) (cos2 x) dx

Solution: We have,
∫ (sin3 x) (cos2 x) dx = ∫ (sin2 x) (cos2 x) (sin x) dx

We know that sin2 x = (1 – cos2 x). we get
∫ (1 – cos2 x) (cos2 x) (sin x) dx

let cos x = t, =>t – sin x dx = dt.

Therefore,
∫ (1 – cos2 x) (cos2 x) (sin x) dx = – ∫ (1 – t2) t2 dt
= – ∫ (t2 – t4) dt
= – [(t3/3) – (t5/5)] + C
Hence, ∫ (sin3 x) (cos2 x) dx = – (1/3) cos3 x + (1/5) cos5 x + C

Integration using trigonometric identities:

  • sin(α+β)=sin(α). cos(β)+cos(α). sin(β)
  • sin(α–β)=sinα. cosβ–cosα. sinβ
  • cos(α+β)=cosα. cosβ–sinα. sinβ
  • cos(α–β)=cosα. cosβ+sinα. sinβ

  • sin 2θ = 2 sinθ cosθ
  • cos 2θ = cos2θ – sinθ = 2 cos2θ – 1 = 1 – 2sin2 θ
  • tan 2θ = (2tanθ)/(1 – tan2θ)
  • sin (θ/2) = ±√[(1 – cosθ)/2]
  • cos (θ/2) = ±√(1 + cosθ)/2
  • tan (θ/2) = ±√[(1 – cosθ)(1 + cosθ)]
  • Sin A. Sin B = [Cos (A – B) – Cos (A + B)]/2
  • Sin A. Cos B = [Sin (A + B) – Sin (A – B)]/2
  • Cos A. Cos B = [Cos (A + B) – Cos (A – B)]/2
  • sin2 x + cos2 x = 1
    1+tan2 x = sec2 x
    cosec2 x = 1 + cot2 x
  • sin2 x =(1-cos2x)/2
  • cos2 x =(1+cos2x)/2
    tan2 x = sec2 x-1
    cosec2 x - 1 = cot2 x
  • sin 3x = 3 sin x – 4 sin3 x,
  • cos 3x = 4 cos3 x -3 cos x  ,

Example :

Solve: ∫sin3x cos2x dx

Solution:

Given: ∫sin3x cos2x dx

∫sin3x cos2x dx = ∫sin2x sin x cos2x dx

We know that sin2x = 1-cos2x.

Therefore,

∫sin3x cos2x dx = ∫(1-cos2x)cos2x sinx dx

Now, let t = cos x, =>  dt = -sin x dx

Hence,

∫sin2x cos2x sin x dx = -∫(1-t2)t2 dt

= -∫(t2-t4) dt

∫sin2x cos2x sin x dx= -[(t3/3) -(t5/5)] + C

Now, let  t = cos x, we get

∫sin2x cos2x sin x dx = -(⅓) cos3x + (⅕)cos5x + C

Therefore, ∫sin3x cos2x dx = -(⅓) cos3x + (⅕)cos5x + C.

Integrals of Some Particular Functions

Integral of function

 

 

we equate the coefficient of x of both the sides to determine the value of A and B, and hence the integral is reduced to one of the known forms.

Example: Find the Integral of the function 

Solution:The given function can be converted into the standard form