Symmetry transformation [closed]

Let $x'=x+\lambda \cos(\omega t)$, where $\omega = \sqrt{k/m}$, be a symmetry transformation of the harmonic oscillator (with mass $m$ and spring constant $k$). My task is now to find a conserved quantity and show that it is actually conserved.

A Hint says to show that the Lagrange function of the harmonic oscillator is "almost" invariant under the transformation.

So the Lagragian for a harmonic oscillator is: $L_0=\frac{1}{2}m\dot{x}^2 - \frac{1}{2}m\omega^2x^2$. If I know put in the transformation I get $$L'=\underbrace{\frac{1}{2}m\dot{x}^2-\frac{1}{2}m\omega^2x^2}_{=L_0} +\underbrace{\left(\frac{1}{2}k\lambda^2-\lambda (m\omega\dot{x}\sin(\omega t)+kx\cos(\omega t))\right)}_{=:\frac{\rm{d}}{\rm{d}t}K_{\lambda}(x,t)},$$ where the idea to realize that the part under the second \underbrace is the total time derivative of a function $K_{\lambda}(x,t)$ came from another hint.

Now I need to find $K_{\lambda}(x,t)$. To be honest that's where the real problem starts: \begin{align}K_{\lambda}(x,t)&=\int\frac{{\rm{d}}}{{\rm{d}}t}K_{\lambda}(x,t){\rm{d}}t\\ &=\int \frac{1}{2}k\lambda^2-\lambda (m\omega\dot{x}\sin(\omega t)+kx\cos(\omega t)) {\rm{d}}t\\ &=\frac{1}{2}k\lambda^2t+C_1-\lambda m\omega\int\dot{x}\sin(\omega t){\rm{d}}t -\lambda k\int x\cos(\omega t){\rm{d}}t \end{align} $\dot{x}$ and $x$ depend on $t$, so how am I supposed to evaluate the integrals?

Thanks for the help in advance, Sito

closed as off-topic by ACuriousMind♦Dec 1 '17 at 13:44

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Of course if you don't know $x$ you cannot evaluate those integrals. However it doesn't mean that you cannot find the function $K$, which depends on $x$, whatever $x$ is. My advice is make $k$ explicit in terms of $m$ and $\omega$, and then stare at the derivative of $K$ long enough to realise that it can be written as a derivative in a simple way.