Daunting Definite Integral

We are given: $\int _{ -1 }^{ 1 }{ x\sqrt { \frac { (2\sin { ({ x }^{ 2 }-1) - \sin { 2({ x }^{ 2 }-1)) } } }{ (2\sin { ({ x }^{ 2 }-1) + \sin { 2({ x }^{ 2 }-1)) } } } } dx }$

*Substitute u = $({ x }^{ 2 }-1)$

= $\int _{ -1 }^{ 1 }{ x\sqrt { \frac { (2\sin { (u)-\sin { 2(u)) } } }{ (2\sin { (u)+ \sin { 2(u)) } } } } dx }$

**if u = $({ x }^{ 2 }-1)$ , $du = 2xdx$ so $\frac { du }{ 2 } =xdx$

= $\int _{ -1 }^{ 1 }{ \frac { 1 }{ 2 } \sqrt { \frac { (2\sin { (u)-\sin { 2(u)) } } }{ (2\sin { (u)+\sin { 2(u)) } } } } du }$

* When defining the bounds of the integral, in terms of u, one must plug the two already given bounds into what u was defined as. This would make the two bounds as 0.

= $\int _{ 0 }^{ 0 }{ \frac { 1 }{ 2 } \sqrt { \frac { (2\sin { (u)-\sin { 2(u)) } } }{ (2\sin { (u)+\sin { 2(u)) } } } } du }$

= $\boxed {0}$

$\begin{aligned} I & = \int_{-1}^1 x \sqrt{\frac {2\sin(x^2-1)-\sin 2(x^2-1)}{2\sin(x^2-1)+\sin 2(x^2-1)}} dx & \small \color{#3D99F6} \text{By }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_{-1}^1 \left( x \sqrt{\frac {2\sin(x^2-1)-\sin 2(x^2-1)}{2\sin(x^2-1)+\sin 2(x^2-1)}} + {\color{#3D99F6}(0-x)} \sqrt{\frac {2\sin({\color{#3D99F6}(0-x)} ^2-1)-\sin 2({\color{#3D99F6}(0-x)} ^2-1)}{2\sin({\color{#3D99F6}(0-x)} ^2-1)+\sin 2({\color{#3D99F6}(0-x)} ^2-1)}} \right) dx \\ & = \frac 12 \int_{-1}^1 \left( x \sqrt{\frac {2\sin(x^2-1)-\sin 2(x^2-1)}{2\sin(x^2-1)+\sin 2(x^2-1)}} - x \sqrt{\frac {2\sin(x^2-1)-\sin 2(x^2-1)}{2\sin(x^2-1)+\sin 2(x^2-1)}} \right) dx \\ & = \frac 12 \int_{-1}^1 0 \ dx = \boxed 0 \end{aligned}$

You can observe that the integrand is an odd function . You may straightaway make use of the fact: $\int_{-a}^{a}{f(x)dx}=0$ when $f(x)$ is an odd function, i.e. in mathematical terms $f(-x)=-f(x)$ .

The longer approach would be to compute the indefinite integral and put in the upper and lower bounds. First, let $u=x^2-1\implies \frac{1}{2}du=xdx$ . Use this information to transform the integral into $\frac{1}{2}\int{\sqrt{\frac{2\sin u -\sin 2u}{2\sin u +\sin 2u}}du}$ . Now make use of the following Trigonometric Identities: $\sin 2x=2\sin x \cos x$ , $\sin^2 x =\frac{1-\cos 2x}{2}$ and $\cos^2 x=\frac{1+\cos 2x}{2}$ . Let $I$ be the definite integral:

$\dfrac{1}{2}\displaystyle \int{\sqrt{\dfrac{1-\cos u}{1+\cos u}}du}=\dfrac{1}{2}\displaystyle \int{\tan \dfrac{u}{2}du}=\log \Biggl| \sec \biggl( \dfrac{x^2-1}{2}\biggr) \Biggr|+C \implies I=\displaystyle \int_{-1}^{1}{x\sqrt{\dfrac{2\sin (x^{2}-1)-\sin 2(x^{2}-1)}{2\sin (x^{2}-1)+\sin 2(x^{2}-1)}}}dx = \log \Biggl| \sec \biggl( \dfrac{x^2-1}{2}\biggr) \Biggr|_{-1}^{1} = 0$

Since every x inside the square root is squared, replacing x for -x will only change the value of the first x before the square root symbol to -x. This means that the area under x=0 to x=1 will be same as the area under x=-1 to x=0, but the area from x=-1 to x=0 will be negative (underneath the x axis) and so will cancel out the area from x=0 to x=1, giving the integral (area above x axis - area under x axis) from x=-1 to x=1 to be 0.