Evaluate: $\int_0^1 \frac{e^x(1+x) \sin^2(x e^x)}{\sin^2(x e^x)+ \sin^2(e-x e^x)} \,dx $

Question

Evaluate: $\int_0^1 \frac{e^x(1+x)\sin^2(x e^x)}{\sin^2(x e^x)+ \sin^2(e-x e^x)} \,dx $

My assumption put $t= x e^x $ then $\int_0^e \frac{\sin^2(t)}{\sin^2(t)+ \sin^2(e-t)} \,dt$ How can I proceed from here? Thanks in advance.

Answer 1

The key with the integral in terms of t is to exploit symmetry by letting $u=e-t$: $$\int_0^e \frac{\sin^2{\left(e-u\right)}}{\sin^2{\left(e-u\right)}+\sin^2{\left(u\right)}} \; du$$

But, $u$ is a dummy variable so you can just switch $u$ to $t$, and add this integral to the integral before the $u$ substitution:

$$2I=\int_0^e \frac{\sin^2{\left(e-t\right)}+\sin^2{\left(t\right)}}{\sin^2{\left(e-t\right)}+\sin^2{\left(t\right)}} \; dt= \int_0^e \; dt=\boxed{\frac{e}{2}}$$