To prove $0.4 \le \int_{0}^{1} x^{\sin x+\cos x} dx \le 0.5$ Prove that  $$0.4 \le \int_{0}^{1} x^{\sin x+\cos x} dx \le 0.5$$
i have proved as follows:
when $x \in (0 \: \: 1)$ $$\sin x+\cos x \in (1 \:\:\:\sin 1+\cos 1)$$
and so $$x^{\sin x+\cos x} \in (x^{\sin 1+\cos 1} \:\:\: x)$$ $\implies$
$$\int_{0}^{1}x^{\sin x+\cos x} \in \left(\int_{0}^{1}x^{\sin 1+\cos 1} \:\:\:\int_{0}^{1} x\right)$$ that is
$$\int_{0}^{1}x^{\sin x+\cos x} \in \left(\frac{1}{\sin 1+\cos 1+1} \: \: \: 0.5\right)$$
But $\frac{1}{\sin 1+\cos 1+1} $ is approximately $0.4$  , is this proof correct?

You may note that $\sin x+\cos x =\sqrt 2\sin(x+\frac \pi 4)$. 
By investigating the monotonicity of $\sin x$, we have, $\forall x \in [0,1]$, $\sin x+\cos x =\sqrt 2\sin(x+\frac \pi 4) \in [1, \sqrt 2]$.
Now can you go further starting from here?
P.S: To get the first identity, we can exploit the sine of the sum identity.

$\sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta$.

First, $a\sin \alpha+b\cos \alpha=\sqrt{a^2+b^2} (\sin \alpha {a\over \sqrt{a^2+b^2}}+\cos \alpha {b\over \sqrt{a^2+b^2}})$.
Obviously there must exist one $\theta$ where $\cos \theta = {a\over \sqrt{a^2+b^2}}$ and $\sin \theta={b\over \sqrt{a^2+b^2}}$, and this leads to $a\sin \alpha+b\cos \alpha=\sqrt{a^2+b^2} (\sin \alpha {a\over \sqrt{a^2+b^2}}+\cos \alpha {b\over \sqrt{a^2+b^2}})=\sqrt{a^2+b^2} (\sin \alpha \cos \theta+\cos \alpha \sin \theta)=\sqrt{a^2+b^2} \sin(\alpha+\theta)$
