Evaluate the following definite integrals :
Problem 1 :
Solution :
1 = A(x-2) + B(x+2)
Put x = 2 1 = B(2+2) 1 = 4B B = 1/4 |
Put x = -2 1 = A(-2-2) 1 = -4A A = -1/4 |
By applying the limits, we get
= (1/4) [ln(2/6) - ln(1/5)]
= (1/4) [ln (2/6) ⋅ (5/1)]
= (1/4) [ln (5/3)]
Problem 2 :
Solution :
By applying the limits, we get
= (1/2) [tan^{-1}(2/2) - tan^{-1}(0)]
= (1/2) [π/4 - 0]
= π/8
Problem 3 :
Solution :
Let us apply trigonometric substitution.
x = cos2θ
(1-x)/(1+x) = (1-cos2θ)/(1+cos2θ)
= sin^{2}θ/cos^{2}θ
= tan^{2}θ
dx = -2sin 2θ dθ
If x = 0 2θ = cos^{-1}x 2θ = cos^{-1}0 2θ = π/2 θ = π/4 |
If x = 1 2θ = cos^{-1}x 2θ = cos^{-1}1 2θ = 0 θ = 0 |
By integrating
= 2[θ-sin2θ/2]
= 2(π/4 - sin2(π/4)/2)
= π/2 - 1
Problem 4 :
Solution :
1+sinx :
sin 2x = 2tanx/(1+tan^{2}x)
Dividing x by 2 on both sides, we get
sin x = 2tan(x/2)/(1+tan^{2}(x/2))
1 + sinx = [1+tan^{2}(x/2) + 2tan(x/2)] / (1+tan^{2}(x/2))
1 + sinx = [1+tan^{2}(x/2) + 2tan(x/2)] / (1+tan^{2}(x/2)) ---(1)
cos 2x = (1-tan^{2}x)/(1+tan^{2}x)
Dividing x by 2 on both sides, we get
cos x = (1-tan^{2}(x/2)/(1+tan^{2}(x/2))
1 + cosx = 1 + [(1-tan^{2}(x/2)/(1+tan^{2}(x/2))]
= [(1+tan^{2}(x/2) + (1-tan^{2}(x/2)] / (1+tan^{2}(x/2))
1 + cosx = 2/(1+tan^{2}(x/2)) ----(2)
(1) / (2)
(1+sinx)/(1+cosx) = [1+tan^{2}(x/2) + 2tan(x/2)]/2
Problem 5 :
Solution :
Let t = cos θ
dt = -sin θ dθ ==> sin θ dθ = -dt
When θ = 0, t = cos 0 ==> 1
When θ = π/2, t = cos π/2 ==> 0
By applying the limit, we get
= (2/3) (1)^{3/2} - (2/7) (1)^{7/2 }- 0
= (2/3) - (2/7)
= (14-6)/21
= 8/21
Problem 6 :
Solution :
By applying the limits, we get
= [sin 2(π/4)]/2
= 1/2
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