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

Question

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

Answer 1

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 !).

Answer 2

Let $x=\sin t$ and $a= \sin^{-1}\frac\pi4$

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

Use the shorthand $t_\pm=\frac 12 (t \pm {\sin t})$ to express the denominator and the numerator as

\begin{align} 1+ \sin t \sin (\sin t)&=1+\frac12[\cos(t-\sin t) - \cos(t+\sin t)]\\ &= \sin^2t_+ + \cos^2t_-\\ \cos ^2t+\cos (\sin t)&=\cos t\cos(t_+ + t_-) + \cos(t_+ - t_-) \\ &= 2 \cos t_- (\sin t_+)’ - 2\sin t_+ (\cos t_-)’ \end{align}

Then

\begin{align} I& =2\int_0^a \frac {\cos t_- d(\sin t_+)-\sin t_+ d(\cos t_-)}{ \sin^2t_+ + \cos^2t_-}\\ & =-2\int_0^a \frac {d\left(\frac{\cos t_-}{\sin t_+}\right) }{ 1 + \left(\frac{\cos t_-}{\sin t_+}\right)^2} =2\cot ^{-1} \frac{\cos \left( \frac a2-\frac\pi8\right) }{\sin \left( \frac a2+\frac\pi8\right) }=1.28553 \end{align}

Answer 3

$$\mathcal{J}=\int_0^{\frac{\pi}{4}}\frac{1+\cos x-x^{2}}{(1+x\sin x)\sqrt {1-x^{2}}}\,dx=\int_0^{\frac{\pi}{4}} \color{red}{\frac{1+x\sin{x}}{\cos{x}\sqrt{1-x^2}}} \cdot \color{blue}{\frac{\cos{x} \left(1+\cos x-x^{2}\right)}{(1+x\sin x)^2}\,dx}$$ If $\displaystyle t=\frac{\sin{x}+x}{1+x\sin{x}} \text{ then }\color{blue}{dt=\frac{\cos{x}\left(1+\cos{x}-x^2\right)}{\left(1+x\sin{x}\right)^2} dx}$, and: \begin{align*} \color{red}{\frac{1+x\sin{x}}{\cos{x}\sqrt{1-x^2}}} &= \frac{1+x\sin{x}}{\sqrt{(1-\sin^2{x})(1-x^2)}}\\ &= \frac{1+x\sin{x}}{\sqrt{\left(1+x\sin{x}\right)^2-\left(\sin{x}+x\right)^2}}\\ &= \frac{1}{\sqrt{1-\left(\frac{\sin{x}+x}{1+x\sin{x}}\right)^2}} \\ &= \color{red}{\frac{1}{\sqrt{1-t^2}} } \end{align*} Putting this together, \begin{align*} \mathcal{J}&=\int_0^{\frac{\pi}{4}} \color{red}{\frac{1+x\sin{x}}{\cos{x}\sqrt{1-x^2}}} \cdot \color{blue}{\frac{\cos{x} \left(1+\cos x-x^{2}\right)}{(1+x\sin x)^2}\,dx} \\&=\int_0^{\frac{4+\pi \sqrt{2}}{\pi + 4\sqrt{2}}} \frac{dt}{\sqrt{1-t^2}} \\ &= \color{green}{\boxed{\arcsin{\left(\frac{4+\pi \sqrt{2}}{\pi + 4\sqrt{2}}\right)}}} \end{align*}