Apparently impossible indefinite integral?

[greg_rack]

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?

 

[PeroK]

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 $u = \sqrt{3 - sin(2x)}$.

 

[PeroK]

Are you sure you have the correct integral?

 

[Haborix]

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.

 

[martinbn]

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?

 

[Frabjous]

Are you sure it is sin2x and not sinx?

 

[George Jones]

Maple gives an answer that is several computer screens long.

 

[aheight]

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.

 

[greg_rack]

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 :)

 

[Mark44]

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.

 

[vela]

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

 

[MathematicalPhysicist]

Maple gives an answer that is several computer screens long.

What an orgasmic integral...
Quite hypnotic!

 

[martinbn]

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?

 

[MathematicalPhysicist]

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.

 

[MathematicalPhysicist]

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

 

[vela]

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
$$\frac{1}{2} \sin x \,\sqrt{3-\sin 2 x}+
\sin ^{-1}\left[\frac{1}{\sqrt 2} \sin \left(x+\frac{\pi }{4}\right)\right]+
\frac 12 \sinh ^{-1}\left[\sin \left(x-\frac{\pi }{4}\right)\right].$$

 

[MathematicalPhysicist]

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?

 

[haruspex]

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?

 

[greg_rack]

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]

View attachment 280496

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

 

[greg_rack]

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!

 

[vela]

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.

 

[MathematicalPhysicist]

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.