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.

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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.

Log in to reply

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.

Log in to reply

What makes you think that it has a closed form?

Log in to reply

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.

Log in to reply

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, $b \ne 1$.

Log in to reply

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.

Log in to reply

You can't solve it by simple Partial Fractions because the denominator of this function does not have any "nice" roots.

Log in to reply

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

Log in to reply

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

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

Log in to reply

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?

Log in to reply

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.

Log in to reply