# Apparently impossible indefinite integral?

• greg_rack
In summary, a student attempted to integrate a function by parts but found that it was impossible. He then tried plotting the function in an integral solver but no antiderivative was found. He then tried solving the problem using integration by parts but got an error.
greg_rack
Gold Member
Homework Statement
$$\int cosx \cdot \sqrt{3-sin(2x)}dx$$
Relevant Equations
none
Hi guys,

I got to solve this integral in a recent test, and literally I had no idea of where to start.
I thought about substituting ##tan(\frac{x}{2})=t## in order to apply trigonometry parametric equations, integrating by parts, substituting, but I always found out I was just running in a circle.
I have even tried to plot the function in an integral solver, but no antiderivative was found.

How's that?

greg_rack said:
Homework Statement:: $$\int cosx \cdot \sqrt{3-sin(2x)}dx$$
Relevant Equations:: none

Hi guys,

I got to solve this integral in a recent test, and literally I had no idea of where to start.
I thought about substituting ##tan(\frac{x}{2})=t## in order to apply trigonometry parametric equations, integrating by parts, substituting, but I always found out I was just running in a circle.
I have even tried to plot the function in an integral solver, but no antiderivative was found.

How's that?
It's doesn't look easy. I'd try ## u = \sqrt{3 - sin(2x)}##.

greg_rack
Are you sure you have the correct integral?

Maybe shift the integration variable by an amount to make the sine function a cosine function in the square root. Then it looks like a half-angle identity might get you somewhere. Haven't tried it, so just a suggestion.

Last edited:
Trying to cheat and putting it in an computer integrator gives me nothing. It could be that it cannot be written in terms of the usual function. May be you mistyped the problem, or was it a definite integral? What is the exact statement of the problem?

greg_rack
Are you sure it is sin2x and not sinx?

Maple gives an answer that is several computer screens long.

Delta2 and greg_rack
It's a relatively short answer (one line of terms) in Mathematica (involving an imaginary term) but I have not been able to solve it by hand although I believe integration by parts is the way to go.

Last edited:
I confirm to y'all that was the actual integral I got in the maths test...
I have tried integrating by parts and substituting as you've said, but still got to nowhere.

Next lesson I'll point out to my teacher that not even PF guys have been able to solve it, hoping she'll be clement with the evaluation :)

greg_rack said:
Next lesson I'll point out to my teacher that not even PF guys have been able to solve it, hoping she'll be clement with the evaluation :)
It's not unheard of for an instructor to come up with a calculus problem that he/she hasn't solved beforehand, and one that is much harder than was intended.

I'd guess it was a typo where the exponent on sine became the coefficient of x.

Vanadium 50, Frabjous, hutchphd and 1 other person
George Jones said:
Maple gives an answer that is several computer screens long.
What an orgasmic integral...
Quite hypnotic!

aheight said:
It's a relatively short answer (one line of terms) in Mathematica (involving an imaginary term) but I have not been able to solve it by hand although I believe integration by parts is the way to go.
Can we see it?

martinbn said:
Can we see it?
I also used Mathematica and I get the following value:
$$\frac{1}{4} \left(2 \sin (x) \sqrt{3-\sin (2 x)}-i \log \left(i \sin (x)+\sqrt{3-\sin (2 x)}+i \cos (x)\right)+3 \tan ^{-1}\left(\frac{\sin (x)+\cos (x)}{\sqrt{3-\sin (2 x)}}\right)-2 \tanh ^{-1}\left(\frac{\cos (x)-\sin (x)}{\sqrt{3-\sin (2 x)}}\right)\right)$$

Interesting integral, surely not for first year calculus 2 course, not even for maths majors first year.

Frabjous
And don't look at what maple spits out.
It's quite magnificent and complicated.

MathematicalPhysicist said:
I also used Mathematica and I get the following value:
$$\frac{1}{4} \left(2 \sin (x) \sqrt{3-\sin (2 x)}-i \log \left(i \sin (x)+\sqrt{3-\sin (2 x)}+i \cos (x)\right)+3 \tan ^{-1}\left(\frac{\sin (x)+\cos (x)}{\sqrt{3-\sin (2 x)}}\right)-2 \tanh ^{-1}\left(\frac{\cos (x)-\sin (x)}{\sqrt{3-\sin (2 x)}}\right)\right)$$
The log term is equal to the arctan plus an imaginary constant, so omitting the constant, I was able to simplify the solution to
$$\frac{1}{2} \sin x \,\sqrt{3-\sin 2 x}+\tan ^{-1}\left(\frac{\sin x+\cos x}{\sqrt{3-\sin 2x}}\right)-\frac{1}{2} \tanh ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{3-\sin 2x}}\right).$$ It also turns out that
\begin{align*}
(3-\sin 2x) + (\sin x + \cos x)^2 &= 4 \\
(3-\sin 2x) - (\sin x - \cos x)^2 &= 2
\end{align*} so one can also express the solution as
$$\frac{1}{2} \sin x \,\sqrt{3-\sin 2 x}+ \sin ^{-1}\left[\frac{1}{\sqrt 2} \sin \left(x+\frac{\pi }{4}\right)\right]+ \frac 12 \sinh ^{-1}\left[\sin \left(x-\frac{\pi }{4}\right)\right].$$

docnet, Vanadium 50, MathematicalPhysicist and 1 other person
vela said:
The log term is equal to the arctan plus an imaginary constant, so omitting the constant, I was able to simplify the solution to
12sin⁡x3−sin⁡2x+tan−1⁡(sin⁡x+cos⁡x3−sin⁡2x)−12tanh−1⁡(sin⁡x−cos⁡x3−sin⁡2x). It also turns out that
\begin{align*}
(3-\sin 2x) + (\sin x + \cos x)^2 &= 4 \
(3-\sin 2x) - (\sin x - \cos x)^2 &= 2
\end{align*} so one can also express the solution as
$$\frac{1}{2} \sin x \,\sqrt{3-\sin 2 x}+ \sin ^{-1}\left[\frac{1}{\sqrt 2} \sin \left(x+\frac{\pi }{4}\right)\right]+ \frac 12 \sinh ^{-1}\left[\sin \left(x-\frac{\pi }{4}\right)\right].$$
Now the big question is how to solve this problem with integration by parts and substitution?

greg_rack said:
I confirm to y'all that was the actual integral I got in the maths test...
I trust that you have checked post#1 is an accurate rendition of how you read the question, but there is still the faint possibility you are misinterpreting it. Could you post an image of the original?

haruspex said:
I trust that you have checked post#1 is an accurate rendition of how you read the question, but there is still the faint possibility you are misinterpreting it. Could you post an image of the original?

greg_rack said:
OK, thanks.
The lack of parentheses is poor, but it clearly is sin 2 not sin2.

haruspex said:
OK, thanks.
The lack of parentheses is poor, but it clearly is sin 2 not sin2.
Yup, I'd guess so...
On Wednesday I'll point it out to my teacher and hopefully the arcane will be unraveled!

MathematicalPhysicist said:
Now the big question is how to solve this problem with integration by parts and substitution?
No integration by parts needed! First, show that ##3 - \sin 2x = 2+2 \cos^2(x+\pi/4)##. Then let ##u=x+\pi/4## and use a few trig identities to get
\begin{align*}
\int \cos x \ \sqrt{3-\sin 2x}\,dx &= \int (\cos u + \sin u) \sqrt{1 + \cos^2 u}\,du \\
&= \int \cos u \sqrt{2 - \sin^2 u}\,du + \int \sin u \sqrt{1 + \cos^2 u}\,du
\end{align*} Each of those integrals can be evaluated with a few substitutions to eventually obtain the answer at the bottom of post 16.

aheight, Frabjous, haruspex and 2 others
vela said:
No integration by parts needed! First, show that ##3 - \sin 2x = 2+2 \cos^2(x+\pi/4)##. Then let ##u=x+\pi/4## and use a few trig identities to get
\begin{align*}
\int \cos x \ \sqrt{3-\sin 2x}\,dx &= \int (\cos u + \sin u) \sqrt{1 + \cos^2 u}\,du \\
&= \int \cos u \sqrt{2 - \sin^2 u}\,du + \int \sin u \sqrt{1 + \cos^2 u}\,du
\end{align*} Each of those integrals can be evaluated with a few substitutions to eventually obtain the answer at the bottom of post 16.
Thanks.
You should really try maple. Though I don't understand the discrepancy between mathematica's and maple's answers.
But I am no software wiz.

## 1. What is an apparently impossible indefinite integral?

An apparently impossible indefinite integral is a mathematical problem that cannot be solved using traditional methods. It is a type of integral that does not have a closed-form solution and requires advanced techniques to solve.

## 2. Why is it called "apparently impossible"?

It is called "apparently impossible" because at first glance, it may seem like there is no solution to the integral. However, with the use of advanced techniques and methods, it is possible to find an approximate solution.

## 3. What are some examples of apparently impossible indefinite integrals?

Some examples of apparently impossible indefinite integrals include the integral of e^(x^2), the integral of sin(x^2), and the integral of ln(x)/x.

## 4. How are apparently impossible indefinite integrals solved?

These integrals are typically solved using advanced techniques such as numerical integration, series expansions, or special functions. These methods allow for an approximate solution to be found, rather than a closed-form solution.

## 5. Why are apparently impossible indefinite integrals important?

These integrals are important because they represent real-world problems that cannot be solved using traditional methods. They also challenge mathematicians to develop new techniques and methods for solving complex problems.

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