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I am getting a nice closed form if the numerator is $\cos 5x + \cos 4x$. 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 $\cos(3x) /\cos(2x) = 1 - 2\cos(2x)$ and reducing the powers of trigonometric functions to 1 shows that we are essentially solving for at least one of $\int \sin(ax) \csc(bx) dx$ , $\int \sin(ax) \sec(bx) dx$, $\int \cos(ax) \sec(bx) dx$, $\int \cos(ax) \csc(bx) dx$ which can't be stated in terms of elementary functions because for all of these cases, $b \ne 1$.

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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## Comments

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TopNewestI am getting a nice closed form if the numerator is $\cos 5x + \cos 4x$. 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 $\cos(3x) /\cos(2x) = 1 - 2\cos(2x)$ and reducing the powers of trigonometric functions to 1 shows that we are essentially solving for at least one of $\int \sin(ax) \csc(bx) dx$ , $\int \sin(ax) \sec(bx) dx$, $\int \cos(ax) \sec(bx) dx$, $\int \cos(ax) \csc(bx) dx$ which can't be stated in terms of elementary functions because for all of these cases, $b \ne 1$.

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I know a method but haven't done it yet.

Write $\cos^5(x)$ and $\cos^4(x)$ as $\frac{1}{16}(10 \cos(x) + 5 \cos(3x) + \cos(5x))$ and $\frac{1}{8}(3 + 4 \cos(2x) + \cos(4x))$.

Substitute $z = {e}^{ix}$.

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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$\frac1{x^3 + 3x^2 + 5x+7}$ have a nice form?

We might have different opinion of "nice" closed form. Does the integration ofLog in to reply

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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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