Solve this integral $\int \frac{dx} {\sin^5{x}}$

Question

Let us denote

$$I_{m,n} = \int \sin^m{x} \cos^n{x}\ dx$$

where $m,n$ are integers (possibly negative or zeros).

There are some well-known recurrent formulas for $I_{m,n}$

So... as an example I was trying to solve this particular integral

$$\int \frac{dx} {\sin^5{x}}$$

using these recurrent formulas and I finally got this answer
(the computations were not very long, just 1 or 2 A4 sheets):

$$F(x) = \frac{5}{8} \ln {|\tan \frac{x}{2}|} - \frac{5}{24} \cdot \frac{\cos{x}}{\sin^2{x}} - \frac{1}{6} \cdot \frac{\cos{x}}{\sin^4{x}}$$

But WA is not giving me a simple expression when I differentiate $F(x)$

Is my answer incorrect? How do I check it with some tool other than WA?

Any ideas how to verify?

EDIT: Now I fixed the issues in my calculations and I am getting this answer.

$$\frac{3}{8}\ln|\tan(\frac{x}{2})|-\frac{3}{8}\frac{\cos(x)}{\sin^{2}(x)}-\frac{1}{4}\frac{\cos(x)}{\sin^{4}(x)}$$

But it still does not match with the WA answer...

WA answer

Which one is correct here?

Have you tried asking WA to evaluate $F'(x)-\frac1{\sin^5x}$? — , Sep 14 '20 at 12:30
Personally I would multiply top and bottom by $\sin x$ and change the variable to $\cos x$ and then do partial fractions. — ancient mathematician, Sep 14 '20 at 12:31
Compare https://math.stackexchange.com/q/258754 for a very similar problem. — Martin R, Sep 14 '20 at 12:34
Try to find the derivate of your $F(x)$, and check the result with Wolfram|Alpha... — Anton Vrdoljak, Sep 14 '20 at 12:35
@MartinR I solved $I_{-3,0}$ in the process of solving $I_{-5,0}$ I am just trying to validate my answer. — peter.petrov, Sep 14 '20 at 12:36
@Gae.S. Trying but not very successfully. So I am still not sure if my answer is correct. — peter.petrov, Sep 14 '20 at 12:39
@Gae.S. Hm... OK, well, if you're sure... I will check my computations 1-2 more times. — peter.petrov, Sep 14 '20 at 12:42
@peter.petrov: Gae.S. used Wolfram. I thought one of the conditions for your question was not using Wolfram? — JonathanZ, Sep 14 '20 at 12:45
To @peter.petrov: Did you used a recursive formula like this one: https://math.stackexchange.com/a/3819841/744799 — Anton Vrdoljak, Sep 14 '20 at 12:47
@AntonVrdoljak My formula has $(m-n+2)$ in the denominator. Yours has $m+n$. Maybe this formula of mine is incorrect and hence the whole solution is messed up. Are you sure of your formula there? — peter.petrov, Sep 14 '20 at 12:59
@AntonVrdoljak Oh!!! My formula is wrong there indeed. Sorry. Thanks a lot. The denominator in your formula pointed me to my problem. Thanks. I will redo my calculations now using the correct formulas. — peter.petrov, Sep 14 '20 at 13:05

Michael Rozenberg · Answer 1 · 2020-09-14T12:33:55.303

3

Use $$\frac{1}{\sin^5x}dx=\frac{\sin{x}}{(1-\cos^2x)^3}dx=-\frac{d(\cos{x})}{(1-\cos^2x)^3}$$

edited Sep 14 '20 at 12:33

answered Sep 14 '20 at 12:31

Michael Rozenberg

194,933

To the downvoter: What did you find wrong here? – 19aksh Sep 14 '20 at 12:33
5

@Ak19 I didn't downvote but this does not answer my question at all. I am asking for verification of answer, this is just throwing some formula and getting upvotes. Not useful, sorry. I will try this approach too but it's another thing I am asking. – peter.petrov Sep 14 '20 at 12:37
@peter.petrov Give another example and I'll show, what we can made. Ask your question. – Michael Rozenberg Sep 14 '20 at 12:43
Well... other folks said my answer is wrong so I will check my calculations again but... I am still not sure if it's wrong or not :) – peter.petrov Sep 14 '20 at 12:44
My questions were there: "Is my answer incorrect? How do I check it with some tool other than WA? Any ideas how to verify?" That's all. – peter.petrov Sep 14 '20 at 12:44
@peter.petrov I would be just calculate the integral. – Michael Rozenberg Sep 14 '20 at 12:47

score 1 · Answer 2 · 2020-09-14T12:49:20.063

1

You rationalize by the simple change of variable $t:=\cos x$:

$$\int\frac{dt}{(1-t^2)^3}.$$

The partial fraction decomposition is

$$\frac3{16(1+t)}-\frac3{16(1-t)}+\frac3{16(1+t)^2}+\frac3{16(1-t)^2}+\frac1{8(1+t)^3}-\frac1{8(1-t)^3}$$ and causes no difficulty.

$$\frac3{16}\log(1-\cos^2x)-\frac3{8(1-\cos^2 x)}+\frac{1+\cos^2x}{8(1-\cos^2 x)^2}.$$

edited Sep 14 '20 at 12:49

answered Sep 14 '20 at 12:37

Integrating rational functions is not yet introduced even (at this point of the book) but OK... As I said the idea was to use these recurrences. Anyway. Thanks. I will check my calculations. – peter.petrov Sep 14 '20 at 12:47
@peter.petrov: the integration of these simple fractions is... simple. – Sep 14 '20 at 12:49
OK... well.. good, thanks – peter.petrov Sep 14 '20 at 12:50
This was the way I also initially thought of, noice. ${(+1)}$ – Riemann'sPointyNose Sep 14 '20 at 19:25

Alessio K · Accepted Answer · 2020-09-14T19:22:46.003

You made a mistake somewhere. From WA and after simplifications you should have (what I get):

$$\frac{3}{8}\ln|\tan(\frac{x}{2})|-\frac{3}{8}\frac{\cos(x)}{\sin^{2}(x)}-\frac{1}{4}\frac{\cos(x)}{\sin^{4}(x)}$$

Let me know!

I think the $24\ln|\tan(\frac{x}{2})|$ is clear and we multiplied it by $\frac{1}{64}$ to give the correct term.

Then re-write the remaining terms (from WA) as:

$$\frac{1}{64}\big[-\frac{1}{\sin^{4}(\frac{x}{2})}-\frac{6}{\sin^{2}({\frac{x}{2})}}+\frac{1}{\cos^{4}(\frac{x}{2})}+\frac{6}{\cos^{2}({\frac{x}{2})}}\big]$$ $$=\frac{6}{64}\big[\frac{\sin^{2}(\frac{x}{2})-\cos^{2}(\frac{x}{2})}{\sin^2({\frac{x}{2})\cos^{2}(\frac{x}{2})}}\big]+\frac{1}{64}\big[\frac{\sin^{4}(\frac{x}{2})-\cos^{4}(\frac{x}{2})}{\sin^4({\frac{x}{2})\cos^{4}(\frac{x}{2})}}\big].$$

Then using $\cos^2(x)-\sin^2(x)=\cos(2x)$ and $\sin(2x)=2\sin(x)\cos(x)$ the first term becomes:

$$\frac{6}{64}\big[\frac{\sin^{2}(\frac{x}{2})-\cos^{2}(\frac{x}{2})}{\sin^2({\frac{x}{2})\cos^{2}(\frac{x}{2})}}\big]=-\frac{3}{32}\big[\frac{\cos(x)}{\frac{\sin^{2}(x)}{4}}\big]=-\frac{3}{8}\frac{\cos(x)}{\sin^{2}(x)}.$$

For the other term we have (numerator is of the form $(A^2-B^2)=(A-B)(A+B)$):

$$-\frac{1}{64}\big[\frac{\cos^{2}(\frac{x}{2})-\sin^2({\frac{x}{2})}}{\frac{\sin^{4}(x)}{2^4}}\big]=-\frac{1}{4}\frac{\cos(x)}{\sin^{4}(x)}.$$

Yeah, I just found the mistake with some help from @AntonVrdoljak — peter.petrov, Sep 14 '20 at 13:06
I get the same answer now but WA gives some very complicated expression. — peter.petrov, Sep 14 '20 at 18:39

score 1 · Answer 4 · answered Sep 14 '20 at 19:01

1

Wolfram Alpha agrees with you. You just typed in the coefficients wrong when specifying the putative antiderivative you wanted differentiated. It took me some experimenting with brackets to get it to understand the corrected query.

answered Sep 14 '20 at 19:01

J.G.

115,835

Phew... OK, indeed. That is a relief, thanks :) – peter.petrov Sep 14 '20 at 19:07

score 0 · Answer 5 · answered Sep 14 '20 at 12:36

I would like to suggest another method: reduction formula for the cosecant (you know that $1$ over sine is the cosecant function):

$$\int \csc^m(x)\ \text{d}x = -\dfrac{\cos(x)\csc^{m-1}(x)}{m-1} + \frac{m-2}{m-1}\int \csc^{m-2}(x)\ \text{d}x$$

Use it twice.

During the process you will have to integrate the cosecant itself, which is a well known integral:

$$\int \csc(x)\ \text{d}x = -\ln(\cot(x) + \csc(x))$$

And eventually the result is:

$$-\frac{1}{64} \csc ^4\left(\frac{x}{2}\right)-\frac{3}{32} \csc ^2\left(\frac{x}{2}\right)+\frac{1}{64} \sec ^4\left(\frac{x}{2}\right)+\frac{3}{32} \sec ^2\left(\frac{x}{2}\right)+\frac{3}{8} \log \left(\sin \left(\frac{x}{2}\right)\right)-\frac{3}{8} \log \left(\cos \left(\frac{x}{2}\right)\right)$$

Solve this integral $\int \frac{dx} {\sin^5{x}}$

5 Answers5