I try to find a solution for :
$\int_{0}^{R}{\frac{x\cos(a+bx^2)}{\sqrt{1+x^2}}dx}$
where $R$ is a strictly positive real and $a$ and $b$ are constant real parameters.
Inspired on this post, I tried a variable change: $x=\sin\theta$. This leads me to a new expression (I'm not sure how to manage boundaries):
$\int^{?}_0{\cos(a+b\sin^2\theta)\sin\theta{}d\theta}$
Now I am stuck (I can't see identification with Anger or Bessel function like in above mentionned post). I have some ideas, but none of them seems to leads to a convenient solution :
- trying an integration by part on this last expression (I am still bothered by the squared sine function into cosine)?
- trying another variable change (I thought of doing the change on $x^2$, but I then lose the simplification in the square root) ?
- is there a known identification for the obtained expression (not in my knowledge)?
- should I try something else than variable change ?
Thank you for your help.
Best Answer
Making the problem more general $$I=\int{\frac{x\cos(a+bx^2)}{\sqrt{c^2+x^2}}dx}$$ let $$u=\sqrt{c^2+x^2}\implies x=\sqrt{u^2-c^2}\implies dx=\frac{u}{\sqrt{u^2-c^2}}du$$ $$I=\int\cos \left(a-b c^2+b u^2\right) \,du$$ Now $v =b u^2$ to make $$I=\frac{1}{\sqrt{b}}\int \cos \left(a-b c^2+v^2\right)\,dv$$ Now, expand the cosine $$I=\frac {\cos(a-b c^2)}{\sqrt{b}}\int \cos(v^2)\,dv-\frac {\sin(a-b c^2)}{\sqrt{b}}\int \sin(v^2)\,dv$$ and, now, we just face Fresnel sine and cosine integrals.
Back to $x$ $$I=\frac{1}{\sqrt{b}}\sqrt{\frac{\pi }{2}}\Bigg[\cos \left(a-b c^2\right) C\left(\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c^2+x^2}\right)-\sin \left(a-b c^2\right) S\left(\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c^2+x^2}\right) \Bigg]$$