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I am getting a nice closed form if the numerator is cos5x+cos4x. I think if it is a JEE problem then the numerator should be what I have written. Can you please check? Otherwise I'll try with this again.
This problem was given to me by my class mate. I tried every integration technique possible. Even I doubt if there exists a valid closed form for it. Please help me sir.
There's no simple closed form without using hypergeometric functions. Apply cos(3x)/cos(2x)=1−2cos(2x) and reducing the powers of trigonometric functions to 1 shows that we are essentially solving for at least one of ∫sin(ax)csc(bx)dx , ∫sin(ax)sec(bx)dx, ∫cos(ax)sec(bx)dx, ∫cos(ax)csc(bx)dx which can't be stated in terms of elementary functions because for all of these cases, b=1.
Oh! I see. I hadn't checked it so I might be wrong. If it doesn't have "nice" roots then also it doesn't matter, computer will do it, it doesn't discriminate b/w real and complex :P
I don't think this is an IITJEE question. This question is not integrable to our knowledge(atleast till JEE point of view). You can use higher level integration to solve this.
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Top NewestI am getting a nice closed form if the numerator is cos5x+cos4x. I think if it is a JEE problem then the numerator should be what I have written. Can you please check? Otherwise I'll try with this again.
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I am sure about the question. The reason I put the jee tag was to get to know if there are any methods of jee applicable.
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What makes you think that it has a closed form?
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This problem was given to me by my class mate. I tried every integration technique possible. Even I doubt if there exists a valid closed form for it. Please help me sir.
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There's no simple closed form without using hypergeometric functions. Apply cos(3x)/cos(2x)=1−2cos(2x) and reducing the powers of trigonometric functions to 1 shows that we are essentially solving for at least one of ∫sin(ax)csc(bx)dx , ∫sin(ax)sec(bx)dx, ∫cos(ax)sec(bx)dx, ∫cos(ax)csc(bx)dx which can't be stated in terms of elementary functions because for all of these cases, b=1.
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I know a method but haven't done it yet.
Write cos5(x) and cos4(x) as 161(10cos(x)+5cos(3x)+cos(5x)) and 81(3+4cos(2x)+cos(4x)).
Substitute z=eix.
Then, you would get a rational polynomial function in terms of z which can "easily" be solved using Partial Fraction or Division approach.
I know this method is way too tedious but that's the most general way to tackle these types of problems.
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You can't solve it by simple Partial Fractions because the denominator of this function does not have any "nice" roots.
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Oh! I see. I hadn't checked it so I might be wrong. If it doesn't have "nice" roots then also it doesn't matter, computer will do it, it doesn't discriminate b/w real and complex :P
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We might have different opinion of "nice" closed form. Does the integration of x3+3x2+5x+71 have a nice form?
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That's quite nice! :P For me, everything has a nice "closed"(I didn't use this word originally) form. Even the error function does! :P
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To me, even the error function is not "nice". I only consider elementary functions to be "nice".
If you consider all of these to have a "nice" closed form, then it's hard to judge whether an integral is worth solving or not. Don't you think so?
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I don't think this is an IITJEE question. This question is not integrable to our knowledge(atleast till JEE point of view). You can use higher level integration to solve this.
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