Evaluate the following integral:
Let us assume
Adding – 1 and + 1
–
Let
Thus I = I1 – I2 …….equation 1
Solving for I1
since
I1 = [tan – 1(∞) – tan – 1(0)]
I1 = π/2 ……….equation 2
Solving for I2
Let .....…..equation 3
a + b = 0; a + c = 1; b + c = 0
solving we get
a = c = 1/2
b = – 1/2
substituting the values in equation 3
Thus substituting the values in I2, thus
Solving :
Let 1 + x2 = y
2xdx = dy
For x = ∞
y = ∞
For x = 0
y = 0
substituting values
Thus
……….equation 4
Substituting values equation 2 and equation 4 in equation 1
Thus
I = I1 – I2
I = π/2 – π/4
I = π/4