11. Some Properties of Definite Integrals and evaluation of definite integrals

Some Properties of Definite Integrals and evaluation of definite integrals

Example :

Prove that ₀∫^π/2 (2log sinx – log sin 2x)dx  = – (π/2) log 2

Solution:

To prove: ₀∫^π/2 (2log sinx – log sin 2x)dx  = – (π/2) log 2 

Proof:

Let I = ₀∫^π/2 (2log sinx – log sin 2x)dx ………………………….…(1)

By using the property of definite integral

₀∫^a f(x) dx = ₀∫^a f(a-x) dx 

Now, apply the property in (1), we get

• •  I = ₀∫^π/2 2log sin[(π/2)-x] – log sin 2[(π/2)-x])dx 

The above expression can be written as

• •  I = ₀∫^π/2 [2log cosx- log sin(π-2x)]dx (Since, sin (90-θ = cos θ)
  •  I = ₀∫^π/2 [2log cosx- log sin2x]dx ……………………………………………………..(2)

Now, add the equation (1) and (2), we get

• • I+ I = ₀∫^π/2 [(2log sinx – log sin 2x) +(2log cosx- log sin2x)]dx
  • 2I =  ₀∫^π/2 [2log sinx -2log 2sinx + 2log cos x]dx
  • 2I = 2 ₀∫^π/2 [log sinx -log 2sinx + log cos x]dx

• •  = ₀∫^π/2 [log sinx + log cos x- log 2sinx]dx

• • I = ₀∫^π/2log[(sinx. cos x)/sin2x]dx

We know that sin2x= 2 sinx cos x)

• • I = ₀∫^π/2log[(sinx. cos x)/(2 sinx cos x)]dx
  • I = ₀∫^π/2 log(1/2)dx
  • I = ₀∫^π/2 (log1-log 2)dx [Since, log (a/b) = log a- log b]
  • I = ₀∫^π/2 -log 2 dx (value of log 1 = 0)

Now, take the constant – log 2 outside the integral,

• • I = -log 2 ₀∫^π/2dx

Now, integrate the function

• • I = -log 2 [x]₀^π/2

Now, substitute the limits

• • I = -log 2 [(π/2)-0]
  • I = – log 2 (π/2)
  • I = – (π/2) log 2 = R.H.S

Therefore, L.H. S = R.H.S

Hence. ₀∫^π/2 (2log sinx – log sin 2x)dx  = – (π/2) log 2 is proved.

Example: solve 

Solution:

Question. solve 

Solution:

Now applying limits, we get,

2I = π/3 – π/6

2I = (2π – π)/6

2I = π/6

I = π/12

Question.  solve 

Solution:

Now applying limits, we get,

2I = π/3 – π/6

2I = (2π – π)/6

2I = π/6

I = π/12

Question.  solve 

Solution:

Question.  solve 

Solution: