Integration of $\sqrt{ x^2 - x\cdot \sin(x) + 4}$

Question

How to integrate

$$\sqrt{ x^2 - x\cdot \sin(x) + 4}$$

with respect to $dx$?

I tried solving this using substitution, but it doesn't seem to work. Can somebody give me a little direction on how to proceed.

Answer 1

This does not appear to have an elementary antiderivative. Maple didn't find one; Wolfram Alpha timed out.

Answer 2

You can try this substitution below. From the point I stopped on, it should be a matter of organizing things and doing basic stuff.

$$\eqalign{ & \int {\sqrt {{x^2} - x\sin x + 4} } dx = \int {\sqrt {4\left( {{{{x^2}} \over 4} - {{x\sin x} \over 4} + 1} \right)} } dx = 2\int {\left( {{{{x^2}} \over 4} - {{x\sin x} \over 4} + 1} \right)} dx \cr & x = 2\tan \theta \cr & {{dx} \over {d\theta }} = 2{\sec ^2}\theta \cr & 2\int {\left( {{{4{{\tan }^2}\theta } \over 4} - {{4ta{n^2}\theta } \over {4\sqrt {4 + 4{{\tan }^2}\theta } }} + 1} \right)} 2{\sec ^2}\theta d\theta \cr & 2\int {\left( {{{\tan }^2}\theta - {{ta{n^2}\theta } \over {2\sec \theta }} + 1} \right)} 2{\sec ^2}\theta d\theta \cr} $$

(sorry if I left anything wrong in the middle of all that!)

Answer 3

Even using special functions, the antiderivative does not seem to be possible.

For the integral, series expansion around $x=0$ could be a possible solution since $$\sqrt{ x^2 - x\, \sin(x) + 4}=2+\frac{x^4}{24}-\frac{x^6}{480}-\frac{31 x^8}{80640}+\frac{31 x^{10}}{725760}+\frac{2213 x^{12}}{319334400}+O\left(x^{14}\right)$$ Trying for $$I_a=\int_0^a\sqrt{ x^2 - x\, \sin(x) + 4}\,dx$$ we would get $$\left( \begin{array}{ccc} a & \text{approximation} & \text{exact} \\ 0.1 & 0.20000 & 0.20000 \\ 0.2 & 0.40000 & 0.40000 \\ 0.3 & 0.60002 & 0.60002 \\ 0.4 & 0.80009 & 0.80009 \\ 0.5 & 1.00026 & 1.00026 \\ 0.6 & 1.20064 & 1.20064 \\ 0.7 & 1.40137 & 1.40137 \\ 0.8 & 1.60266 & 1.60266 \\ 0.9 & 1.80476 & 1.80476 \\ 1.0 & 2.00800 & 2.00800 \\ 1.1 & 2.21275 & 2.21275 \\ 1.2 & 2.41948 & 2.41948 \\ 1.3 & 2.62871 & 2.62870 \\ 1.4 & 2.84100 & 2.84098 \\ 1.5 & 3.05699 & 3.05695 \\ 1.6 & 3.27738 & 3.27727 \\ 1.7 & 3.50290 & 3.50264 \\ 1.8 & 3.73438 & 3.73377 \\ 1.9 & 3.97272 & 3.97137 \\ 2.0 & 4.21902 & 4.21616 \end{array} \right)$$

Edit

The above approximation is not too bad for "small" values of $a$. We can make if better if we consider $$J_{k,a}=\int_{2k \pi}^{2k\pi+a}\sqrt{ x^2 - x\, \sin(x) + 4}\,dx$$ Chanking variable $x=2k \pi+y$ $$J_{k,a}=\int_{0}^{a}\sqrt{4 \pi ^2 k^2+4 \pi k y-(2 \pi k+y) \sin (y)+y^2+4}\,dy$$ and using the Taylor expansion of $\sin(y)$, the integrand becomes $$2 \sqrt{\pi ^2 k^2+1}\left(1+\frac{\pi k y}{4 (\pi ^2 k^2+1)}-\frac{\left(\pi ^2 k^2\right) y^2}{32 \left(\pi ^2 k^2+1\right)^2}+\frac{\pi k \left(16 \pi ^4 k^4+35 \pi ^2 k^2+16\right) y^3}{384 \left(\pi ^2 k^2+1\right)^3}+\frac{\left(64 \pi ^6 k^6+241 \pi ^4 k^4+320 \pi ^2 k^2+128\right) y^4}{6144 \left(\pi ^2 k^2+1\right)^4}\right)+O\left(y^5\right)$$ for acceptable results as show below $$\left( \begin{array}{cccc} k & a &\text{approximation} & \text{exact} \\ 1 & 0.25 & 1.66333 & 1.66333 \\ 1 & 0.50 & 3.35705 & 3.35704 \\ 1 & 0.75 & 5.08364 & 5.08350 \\ 1 & 1.00 & 6.84782 & 6.84701 \\ 1 & 1.25 & 8.65666 & 8.65350 \\ 1 & 1.50 & 10.5197 & 10.5102 \\ 1 & 1.75 & 12.4494 & 12.4250 \\ 1 & 2.00 & 14.4607 & 14.4060 \\ & \\ 2 & 0.25 & 3.19659 & 3.19659 \\ 2 & 0.50 & 6.42490 & 6.42489 \\ 2 & 0.75 & 9.68763 & 9.68749 \\ 2 & 1.00 & 12.9896 & 12.9888 \\ 2 & 1.25 & 16.3377 & 16.3348 \\ 2 & 1.50 & 19.7412 & 19.7326 \\ 2 & 1.75 & 23.2117 & 23.1901 \\ 2 & 2.00 & 26.7631 & 26.7151 \\ & \\ 3 & 0.25 & 4.75443 & 4.75443 \\ 3 & 0.50 & 9.54087 & 9.54086 \\ 3 & 0.75 & 14.3621 & 14.3620 \\ 3 & 1.00 & 19.2230 & 19.2223 \\ 3 & 1.25 & 24.1304 & 24.1277 \\ 3 & 1.50 & 29.0935 & 29.0853 \\ 3 & 1.75 & 34.1236 & 34.1029 \\ 3 & 2.00 & 39.2341 & 39.1883 \end{array} \right)$$