Integration of this trigonometry function

songoku
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Homework Statement
Find
$$\int x \sqrt{1+4 \cos^2 (x)} dx$$
Relevant Equations
High School Integration:
integration by substitution
integration by part
integration of trigonometry function
Is it possible to do the integration? That is the full question

I don't know where to start, try to use ##u=\cos x## and also ##\cos^2 (x) = \frac{1}{2} + \frac{1}{2} \cos (2x)## but failed.

Thanks
 
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WolframAlpha and I assume that there is no closed solution. The fact that ##x## occurs as polynomial and as trigonometric function (whose derivative is not the polynomial in ##x##) makes it impossible to use things like e.g. the Weierstraß substitution.
 
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Thank you very much fresh_42
 
Are you 100% sure you wrote the problem down right? If instead of x in the cosine you have an x2 you could make more progress. (How much progress would depend on what the new integral is)
 
Vanadium 50 said:
Are you 100% sure you wrote the problem down right? If instead of x in the cosine you have an x2 you could make more progress. (How much progress would depend on what the new integral is)
Yes 100% sure. I have re-checked several times when doing the question and before posting it here
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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