Find :

$$\int_0^{\frac{\pi}{4}}\frac{1+\cos x-x^{2}}{(1+x\sin x)\sqrt {1-x^{2}}}\,dx$$

I don't know how I starte & evaluate this integral

Wolfram alpha give $=1,28553$

My problem whene I use $t=\cos x$ I get $\arccos x$

Same problem with $t=\sin x$

If any one have idea please help me

  • $\begingroup$ Don't see any particular reason for a closed form. $\endgroup$ – Simply Beautiful Art Jul 18 '19 at 1:33

As Simply Beautiful Art commented, it would be very surprising to find a closed form for this integral.

So, either numerical integration or series expansions built at $x=0$. The latest would give $$\frac{1+\cos x-x^{2}}{(1+x\sin x)\sqrt {1-x^{2}}}=2-\frac{5 x^2}{2}+\frac{23 x^4}{8}-\frac{2323 x^6}{720}+\frac{17113 x^8}{4480}-\frac{5186177 x^{10}}{1209600}+O\left(x^{12}\right)$$ which would lead to something which evaluates as $1.27486$ (which is not very good).

Pushing the expansion up to $O\left(x^{24}\right)$ would give $1.28486$ (a lot of effort for a difference of 0.01 only !).

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