Energy-momentum tensor from the variation of action of RNS strings

In exercise 4.6 p. 121 of Becker, Becker, Schwarz's book 'String theory and M-theory', they state that under using the variation $\delta_+X=a^+\partial_+X$ and $\delta_+\psi_A=a^+\partial_+\psi_A$ where $A=\pm$, we may identify the components of the energy momentum tensor of the RNS strings from the variation of the action $$\delta_+S=\frac{1}{\pi}\int d^2\sigma \: \delta_+\mathcal{L},$$ where \begin{align} \delta_+\mathcal{L}&=\delta_+(2\partial_+X\cdot\partial_-X+i\psi_-\cdot\partial_+\psi_-+i\psi_+\cdot\partial_-\psi_+)\\ &=a^+(-2\partial_-(\partial_+X\cdot\partial_+X)+i\partial_+(\psi_+\cdot\partial_-\psi_+)-i\partial_-(\psi_+\cdot\partial_+\psi_+))\\ &=-2a^+(\partial_-T_{++}+\partial_+T_{-+}), \end{align} and similarly using $\delta_-$. So my question is how did the authors go from the first line to the second one, because naively I would think that: \begin{align} \delta_+\mathcal{L}=&\delta_+(2\partial_+X\cdot\partial_-X+i\psi_-\cdot\partial_+\psi_-+i\psi_+\cdot\partial_-\psi_+)\\ =&2[\partial_+(\delta_+X)\cdot\partial_-X+\partial_+X\cdot\partial_-(\delta_+X)]+i[(\delta_+\psi_-)\cdot\partial_+\psi_-+\psi_-\cdot\partial_+(\delta_+\psi_-)]\\ &+i[(\delta_+\psi_+)\cdot\partial_-\psi_++\psi_+\cdot\partial_-(\delta_+\psi_+)]\\ =&a^+\partial_+(2\partial_+X\cdot\partial_-X+i\psi_-\cdot\partial_+\psi_-+i\psi_+\cdot\partial_-\psi_+). \end{align} Then I would find totally incorrect $T_{-+}$ and $T_{++}$.

I will just write the variation for Bosonic field, similar logic will follow for $\psi$.
Start with $2 nd$ line of your calculation and substitute the value of $\delta_+ X$. Expression will look like this $$2 a^+ (\partial_-X) \partial_+ \partial_+ X + 2 a^+ (\partial_+X) \partial_- \partial_+ X$$ Do the integration by parts and throw away boundary terms. After integrating $$-2 \partial_+(\partial_-X) \partial_+X -2 \partial_-(\partial_+X) \partial_+X$$ Using the fact that $\partial_-\partial_+X= \partial_+\partial_-X$, and just take $\partial_-$ comman to full expression. Hence you will get the correct $T_{++}$.