# Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{sech}\,2x = \frac{2}{e^{2x}+e^{-2x}}$ | B1 | For sech$2x$ expression or equivalent |
| $u = e^{2x} \Rightarrow du = 2e^{2x}dx$ **or** $x = \frac{1}{2}\ln u \Rightarrow dx = \frac{1}{2u}du$ | M1 | For differentiating substitution correctly and substituting into *their* integral |
| $\Rightarrow I = \int\frac{2}{(e^{2x}+e^{-2x})}\cdot\frac{du}{2e^{2x}}$ | A1 | For correct integral |
| $= \int\frac{1}{u^2+1}\,du$ | | |
| $= \tan^{-1}u\,(+c) = \tan^{-1}(e^{2x})+c$ | M1 | For integration to $\tan^{-1}(\,)$ |
| | A1 | For correct expression ($c$ required) |

**[5 marks]**

---