Відео

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歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

呢個係core 定m1/2? 我記得core 都有

I did it the exact same way as you did ❤

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

其實真係只有根據個答案倒轉嚟做先work. 難就不太難但感覺唔係"真數學"😢😢😢

講中文ok 嗎

Actually after the first application of lhopital in part b you can kill off the 2nd limit by considering sin(sinx)/2x = (sin(sinx)/sinx)(sinx/x)(1/2) to avoid the 2nd application of the lhopital rule. Also in part a I think it is easier to apply the estimation lemma on abs(the integral) and show that it's 0 by bounding it by abs(x(1-x)) for small enough x. Your method still works perfectly for both parts though.

👍

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(a) ln(1+x)= x-x^2+x^3-x^4+… not 1-x+x^2-x^3+…

Dr Liu, the email I attached a text in which you can find the whole question has sent to you. Thank you!

sorry Dr liu, i lost your email, please help send to me again to send you the whole question

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

thanks for the help

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

可否幫忙。reduction formula for integration from 0 to pi, of cos^2 nx/cos^x😢😢

Hi! Here's an alternative to this problem, which is also workable although it is slower: Still integration by parts, but I will not do it directly from the start. Let u = arccos x, cos u = x So dx/du = - sin u, and dx = - sin u du The integrand then becomes -u² sin u and after performing by parts and undoing the substitution, the answer will be the same

I think this is out of scope in dse, as dse m2 do not require limit of sequences

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

I liked that first integral. A nice problem.

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

Hi! Nice solutions on both questions, but I've come up with some alternatives to them! For Question 1: Alternative method 1: If a student doesn't have those formulae (printed on the M2 exam paper) in hand and forgets this trick and those formulae, this alternative can help them solve an integral of a product of any powers of sin x and cos x, although it is obviously slower: All we need to memorize is e^ix = cos x + i sin x Change x to "-x" yields e^(-ix) = cos x - i sin x Adding the 2 formulae above yields cos x = [e^ix + e^(-ix)]/2 Subtracting the 2 formulae yields sin x = [e^ix - e^(-ix)]/(2i) So sin⁵x cos²x will become: {[e^ix - e^(-ix)]/(2i)}⁵ • {[e^ix + e^(-ix)]/2}² = 1/(2⁷i) • [e^(5ix) - 5e^(3ix) + 10e^ix - 10e^(-ix) + 5e^(-3ix) - e^(-5ix)][e^(2ix) + e^(-2ix) + 2] (By binomial expansion) = 1/(2⁷i) • [e^(5ix) - e^(-5ix) - 5(e^(3ix) - e^(-3ix)) + 10(e^ix - e^(-ix))][e^(2ix) + e^(-2ix) + 2] = 1/(2⁷i) • {2[e^(5ix) - e^(-5ix) - 5(e^(3ix) - e^(-3ix)) + 10(e^ix - e^(-ix))] + e^(7ix) - e^(-3ix) - 5(e^(5ix) - e^(-ix)) + 10(e^(3ix) - e^ix) + e^(3ix) - e^(-7ix) - 5(e^ix - e^(-5ix)) + 10(e^(-ix) - e^(-3ix))} = 1/(2⁷i) • {e^(7ix) - e^(-7ix) + 2[e^(5ix) - e^(-5ix)] - 10[e^(3ix) - e^(-3ix)] + 20[e^ix - e^(-ix)] + e^(3ix) - e^(-3ix) - 5[e^(5ix) - e^(-ix) + e^ix - e^(-5ix)] + 10[e^(3ix) - e^ix + e^(-ix) - e^(-3ix)]} = 1/2⁶ • [sin 7x + 2sin 5x - 10sin 3x + 20sin x + sin 3x - 5(sin 5x + sin x) + 10(sin 3x - sin x)] (Using sin x = [e^ix - e^(-ix)]/(2i)) = 1/2⁶ • (sin 7x - 3sin 5x + sin 3x + 5sin x) So, the integral will become: 1/2⁶ • (-1/7 • cos 7x + 3/5 • cos 5x - 1/3 • cos 3x - 5cos x) + C = -1/448 • cos 7x + 3/320 • cos 5x - 1/192 • cos 3x - 5/64 • cos x + C

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!

stirling

歡迎各位同學, 老師, 家長發表有意義的意見, 多多討論!!!