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junior college 1 | H2 Maths
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This is a J1 integration question. I have difficulty solving part ii of the question :( I tried to dt/dT but I got sec^2 and I’m not sure how to get rid of it. Unless I use by parts??? But I don’t think that’s correct :/
Date Posted: 3 years ago
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t = tan θ/2
t + 1 = tan θ/2 + 1
sin θ = 2t/(1+t²)
sinθ + 1 = 2t/(1+t²) + 1
= (2t+1+t²) / (1+t²)
= (1+t)² / (1+t²)
So, (tan θ/2 + 1) / (sinθ + 1)
= (t+1) / ((1+t)²/(1+t²))
= (1+t) × (1+t²) / (1+t)²
= (1+t²)/(1+t)
Now change the bounds.
When θ = 0, t = tan (0/2) = tan 0 = 0
When θ = π/2, t = tan (π/2)/2 = tan π/4 = 1
So your new bounds are 0,1
Next, you should be doing dθ/dt instead of dt/dθ
t = tan θ/2
So θ/2 = tan-¹ t
θ = 2 tan-¹ t
dθ/dt = 2 (1/(1+t²)) = 2/(1+t²)
(Using the derivative of arctan x / tan-¹ x)
Lastly, putting them altogether,
∫ (tan θ/2 + 1) / (sinθ + 1) dθ
= ∫ ( (1+t²)/(1+t) (2/(1+t²) ) dt
= ∫ 2/(1+t) dt
= [2 ln t]
= [2 ln (1+1)] - [2 ln (1+0)]
= 2 ln 2 - 2 ln 1
= 2 ln 2
(or ln 4)
t + 1 = tan θ/2 + 1
sin θ = 2t/(1+t²)
sinθ + 1 = 2t/(1+t²) + 1
= (2t+1+t²) / (1+t²)
= (1+t)² / (1+t²)
So, (tan θ/2 + 1) / (sinθ + 1)
= (t+1) / ((1+t)²/(1+t²))
= (1+t) × (1+t²) / (1+t)²
= (1+t²)/(1+t)
Now change the bounds.
When θ = 0, t = tan (0/2) = tan 0 = 0
When θ = π/2, t = tan (π/2)/2 = tan π/4 = 1
So your new bounds are 0,1
Next, you should be doing dθ/dt instead of dt/dθ
t = tan θ/2
So θ/2 = tan-¹ t
θ = 2 tan-¹ t
dθ/dt = 2 (1/(1+t²)) = 2/(1+t²)
(Using the derivative of arctan x / tan-¹ x)
Lastly, putting them altogether,
∫ (tan θ/2 + 1) / (sinθ + 1) dθ
= ∫ ( (1+t²)/(1+t) (2/(1+t²) ) dt
= ∫ 2/(1+t) dt
= [2 ln t]
= [2 ln (1+1)] - [2 ln (1+0)]
= 2 ln 2 - 2 ln 1
= 2 ln 2
(or ln 4)
Just a note :
dt/dθ can be done as well, but with some manipulation required.
t = tan θ/2
dt/dθ = ½ sec² θ/2
= ½ (1 + tan² θ/2)
[recall that 1 + tan²x = sec² x]
= ½ (1 + t²) or (1 + t²)/2
So dθ/dt = 2/(1 + t²)
(Recall that the result of dθ/dt and result of dt/dθ are reciprocals of each other.
dθ/dt × dt/dθ = 1 )
So eventually, you will still get the substitution of dθ with 2/(1+ t²) dt
dt/dθ can be done as well, but with some manipulation required.
t = tan θ/2
dt/dθ = ½ sec² θ/2
= ½ (1 + tan² θ/2)
[recall that 1 + tan²x = sec² x]
= ½ (1 + t²) or (1 + t²)/2
So dθ/dt = 2/(1 + t²)
(Recall that the result of dθ/dt and result of dt/dθ are reciprocals of each other.
dθ/dt × dt/dθ = 1 )
So eventually, you will still get the substitution of dθ with 2/(1+ t²) dt
Just realised a typo in the working.
It should be [2 ln|1+t|] instead of [2 ln t]
It should be [2 ln|1+t|] instead of [2 ln t]
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