Hello I am trying to show that $\frac{2}{5} \le \int_0^1 x^{\cos x+\sin x}dx \le \frac{1}{2}$ or to get better bounds.
Here is my try:$$I=\int_0^1 x^{\cos x}\cdot x^{\sin x} \,dx$$ By Cauchy inequality $$I^2\le \int_0^1 x^{\cos x} \,dx \cdot \int_0^1 x^{\sin x} \,dx$$ for $x\in [0,1]\,$ $$ 0\le \cos x \le 1$$ $$\,0\le \sin x \le 1$$ thus $$x^{\cos x} \le x$$ $$x^{\sin x} \le x$$ so $$\int_0^1 x^{\cos x} \,dx \le \frac{1}{2}$$ $$\int_0^1 x^{\sin x} \,dx \le \frac{1}{2}$$ Therefore $$I\le \sqrt{\frac{1}{2}\cdot \frac{1}{2}} \le \frac{1}{2}$$ Is this correct? And what about the lower bound?
Let $f(x)=\cos x+\sin x,\ x\in[0,1]$. $f$ is continuous function on $[0,1]$, so it has maximum and minimum value. It is easy to see that $$1\leq f(x)\leq \sqrt{2}.$$ So $$\frac{2}{5}<\sqrt{2}-1=\int_0^1 x^{\sqrt{2}}\ dx \leq\int_0^1 x^{\cos x+\sin x} \,dx\leq\int_0^1 x\ dx=\frac{1}{2}.$$