Insane Reciprocal Trig Integral
∫ 0 4 π ( cos 2 0 2 0 ( x ) sin ( x ) ) 2 0 2 1 2 d x = ?
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1 solution
Need not to use large font all the way, making it like a kindergarten textbook.
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This is deliberate, since the powers look too small otherwise.
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Not recommended, you don't see it in other problems. I am moderator and I edit problem statements. I have maintained certain standard. For small power purpose. I would use up to \large not \Large. I can actually edit your solution. But of course that would be rude.
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@Chew-Seong Cheong – I have edited my solution as per your recommendation.
@Chew-Seong Cheong – You could edit my solution if you see scope for further improvement. I can incorporate some of the changes for future solutions.
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@Karan Chatrath – Done. This is my style. Only a few words.
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@Chew-Seong Cheong – Thank you!
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@Karan Chatrath – I am relearning Physics. Any online free reference can I use?
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@Chew-Seong Cheong – I would recommend the Feynman lectures. This is reading material which is freely available and very popular.
https://www.feynmanlectures.caltech.edu/
I would also recommend 'Khan Academy' for insightful video explanations. Videos can also be found on Youtube.
There are also several video lectures by physicists such as Richard Feynman, Leonard Susskind, and many others that can be found on Youtube.
Nice problem.
Nitpick: you are using the exponents inconsistently with regard to the trig functions. By which I mean you have
cos
m
u
∗
sin
u
n
. This makes it look like you are mixing the order of exponentiation with regard to the trig function calls. But I don't think that's what you want. You don't want to exponentiate before the trig function call, but that's what you've written. For example,
tan
x
2
0
2
1
2
should be
tan
2
0
2
1
2
x
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Sorry, I edited the solution, it were typos. That was why some are correct and others are not. Thanks for the feed back.
Thanks for the feedback!
I = ∫ 0 4 π ( cos 2 0 2 0 x sin x ) 2 0 2 1 2 d x = ∫ 0 4 π cos 2 0 2 1 4 0 4 0 x sin 2 0 2 1 2 x d x = ∫ 0 4 π cos 2 x tan 2 0 2 1 2 x d x = ∫ 0 4 π tan 2 0 2 1 2 x sec 2 x d x = ∫ 0 1 t − 2 0 2 1 2 d t = 2 0 1 9 2 0 2 1 t 2 0 2 1 2 0 1 9 ∣ ∣ ∣ ∣ 0 1 = 2 0 1 9 2 0 2 1 × cos 2 0 2 1 2 x cos 2 0 2 1 2 x Let t = tan x ⟹ d t = sec 2 x d x