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secondary 4 | A Maths
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How to do this?
Date Posted: 3 years ago
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done
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clear
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Jiayou
Date Posted:
3 years ago
This method is probably too difficult for a Sec 4 student who probably only learns the most basic of integration stuffs (integration by substitution is only seen after Sec 4 during poly or JC studies)
done
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This question is extremely difficult for a Sec 4 student, and I don’t expect this to appear in the O Levels at all.
Date Posted:
3 years ago
Not that difficult actually. Just a bit further of thinking.
Given the increasing occurrence of higher order questions in the past 10 years, I wouldn't be surprised if Cambridge threw out a similar one this year.
Given the increasing occurrence of higher order questions in the past 10 years, I wouldn't be surprised if Cambridge threw out a similar one this year.
Hi Rei, here's an alternative method : (if you're not required to use i)'s result or if you were given this expression to integrate from the start)
∫ (e²ˣ / (e²ˣ + e⁻²ˣ)) dx
= ∫ (e²ˣ / (e²ˣ + e⁻²ˣ) × e²ˣ/e²ˣ) dx
= ∫ (e²ˣ(e²ˣ) / (e²ˣ(e²ˣ) + e²ˣ(e⁻²ˣ)) ) dx
= ∫ (e⁴ˣ/ (e⁴ˣ + 1)) dx
= ¼ ∫ (4e⁴ˣ/ (e⁴ˣ + 1)) dx
= ¼ ln (e⁴ˣ + 1) + c
∫ (e²ˣ / (e²ˣ + e⁻²ˣ)) dx
= ∫ (e²ˣ / (e²ˣ + e⁻²ˣ) × e²ˣ/e²ˣ) dx
= ∫ (e²ˣ(e²ˣ) / (e²ˣ(e²ˣ) + e²ˣ(e⁻²ˣ)) ) dx
= ∫ (e⁴ˣ/ (e⁴ˣ + 1)) dx
= ¼ ∫ (4e⁴ˣ/ (e⁴ˣ + 1)) dx
= ¼ ln (e⁴ˣ + 1) + c
Alternative presentation for the first method :
d/dx ln(e²ˣ + e⁻²ˣ) = (2e²ˣ - 2e⁻²ˣ))/(e²ˣ + e⁻²ˣ)
d/dx ln(e²ˣ + e⁻²ˣ) + 2 = (2e²ˣ - 2e⁻²ˣ)/(e²ˣ + e⁻²ˣ) + 2
d/dx ln(e²ˣ + e⁻²ˣ) + 2 = (2e²ˣ - 2e⁻²ˣ)/(e²ˣ + e⁻²ˣ) + (2e²ˣ + 2e⁻²ˣ)/(e²ˣ + e⁻²ˣ)
d/dx ln(e²ˣ + e⁻²ˣ) + 2 = 4e²ˣ/(e²ˣ + e⁻²ˣ)
∫ (d/dx ln(e²ˣ + e⁻²ˣ) + 2) dx = ∫4e²ˣ/(e²ˣ + e⁻²ˣ) dx
∫ (d/dx ln(e²ˣ + e⁻²ˣ)) dx + ∫ 2 dx = 4∫ e²ˣ/(e²ˣ + e⁻²ˣ) dx
ln(e²ˣ + e⁻²ˣ) + 2x = 4∫ e²ˣ/(e²ˣ + e⁻²ˣ) dx
¼(ln(e²ˣ + e⁻²ˣ) + 2x) = ∫e²ˣ/(e²ˣ + e⁻²ˣ) dx
This can be simplified further into a single term.
¼(ln(e²ˣ + e⁻²ˣ) + 2x ln e) = ∫e²ˣ/(e²ˣ + e⁻²ˣ) dx
¼(ln(e²ˣ + e⁻²ˣ) + ln e²ˣ) = ∫e²ˣ/(e²ˣ + e⁻²ˣ) dx
¼ln((e²ˣ + e⁻²ˣ)e²ˣ) = ∫e²ˣ/(e²ˣ + e⁻²ˣ) dx
¼ln(e⁴ˣ + 1) + c = ∫e²ˣ/(e²ˣ + e⁻²ˣ) dx
(c is a arbitrary constant)
d/dx ln(e²ˣ + e⁻²ˣ) = (2e²ˣ - 2e⁻²ˣ))/(e²ˣ + e⁻²ˣ)
d/dx ln(e²ˣ + e⁻²ˣ) + 2 = (2e²ˣ - 2e⁻²ˣ)/(e²ˣ + e⁻²ˣ) + 2
d/dx ln(e²ˣ + e⁻²ˣ) + 2 = (2e²ˣ - 2e⁻²ˣ)/(e²ˣ + e⁻²ˣ) + (2e²ˣ + 2e⁻²ˣ)/(e²ˣ + e⁻²ˣ)
d/dx ln(e²ˣ + e⁻²ˣ) + 2 = 4e²ˣ/(e²ˣ + e⁻²ˣ)
∫ (d/dx ln(e²ˣ + e⁻²ˣ) + 2) dx = ∫4e²ˣ/(e²ˣ + e⁻²ˣ) dx
∫ (d/dx ln(e²ˣ + e⁻²ˣ)) dx + ∫ 2 dx = 4∫ e²ˣ/(e²ˣ + e⁻²ˣ) dx
ln(e²ˣ + e⁻²ˣ) + 2x = 4∫ e²ˣ/(e²ˣ + e⁻²ˣ) dx
¼(ln(e²ˣ + e⁻²ˣ) + 2x) = ∫e²ˣ/(e²ˣ + e⁻²ˣ) dx
This can be simplified further into a single term.
¼(ln(e²ˣ + e⁻²ˣ) + 2x ln e) = ∫e²ˣ/(e²ˣ + e⁻²ˣ) dx
¼(ln(e²ˣ + e⁻²ˣ) + ln e²ˣ) = ∫e²ˣ/(e²ˣ + e⁻²ˣ) dx
¼ln((e²ˣ + e⁻²ˣ)e²ˣ) = ∫e²ˣ/(e²ˣ + e⁻²ˣ) dx
¼ln(e⁴ˣ + 1) + c = ∫e²ˣ/(e²ˣ + e⁻²ˣ) dx
(c is a arbitrary constant)
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