# Integrals involving roots

I am bit stucked with an integration form while doing one of my proofs for a graphics application.Issue is I cant take out the terms from the trigonometric functions for a proper known integral format. Could you give me some suggestions

Problem 1

$\displaystyle\int\frac{\sin^2\left( \sqrt{ ax^2+bx+c}\right) }{ \sqrt{ ax^2+bx+c}} \operatorname{d}x \tag1$

Problem 2

$\displaystyle\int\frac{\sin \left(2\times \sqrt{ ax^2+bx+c}\right) }{ \sqrt{ ax^2+bx+c}} \operatorname dx \tag2$

• A search on Wolfram Alpha wasn't able to find a closed form solution. – user28375028 Sep 15 '14 at 2:55
• It has no direct solution. I feel we need to do some manipulation before applying knowing known integrals – Nirvana Sep 15 '14 at 2:57

Hint:

For $\int\dfrac{\sin^2\sqrt{ax^2+bx+c}}{\sqrt{ax^2+bx+c}}dx$ ,

$\int\dfrac{\sin^2\sqrt{ax^2+bx+c}}{\sqrt{ax^2+bx+c}}dx$

$=\int\dfrac{1-\cos\left(2\sqrt{ax^2+bx+c}\right)}{2\sqrt{ax^2+bx+c}}dx$

$=-\int\dfrac{1}{2\sqrt{ax^2+bx+c}}\sum\limits_{n=1}^\infty\dfrac{(-1)^n4^n(ax^2+bx+c)^n}{(2n)!}dx$

$=\int\sum\limits_{n=1}^\infty\dfrac{(-1)^{n+1}2^{2n-1}(ax^2+bx+c)^{n-\frac{1}{2}}}{(2n)!}dx$

For $\int\dfrac{\sin\left(2\sqrt{ax^2+bx+c}\right)}{\sqrt{ax^2+bx+c}}dx$ ,

$\int\dfrac{\sin\left(2\sqrt{ax^2+bx+c}\right)}{\sqrt{ax^2+bx+c}}dx$

$=\int\dfrac{1}{\sqrt{ax^2+bx+c}}\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n+1}(ax^2+bx+c)^{n+\frac{1}{2}}}{(2n+1)!}dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n+1}(ax^2+bx+c)^n}{(2n+1)!}dx$

I do not think taht any closed form could be found for these integrals except in the case where $b^2-4ac=0$ where we should find $$a x^2+bx+x=a \Big(x+\frac{b}{2a}\Big)^2$$ $$\sqrt{a x^2+bx+x}=\sqrt a (x+\frac{b}{2a})$$ For such a case $$\displaystyle\int\frac{\sin^2\left( \sqrt{ ax^2+bx+c}\right) }{ \sqrt{ ax^2+bx+c}} dx =\frac{\log \left(\frac{2 a x+b}{2 \sqrt{a}}\right)-\text{Ci}\left(\frac{b+2 a x}{\sqrt{a}}\right)}{2 \sqrt{a}}$$ $$\displaystyle\int\frac{\sin \left(2\times \sqrt{ ax^2+bx+c}\right) }{ \sqrt{ ax^2+bx+c}} dx=\frac{\text{Si}\left(\frac{b+2 a x}{\sqrt{a}}\right)}{\sqrt{a}}$$