June 2025 Paper 1 Q17
17
(a) Use the substitution \(u = \mathrm{e}^{x} + 1\) to show that\[\int \frac{\mathrm{e}^{2x}}{\mathrm{e}^{x} + 1}\,\mathrm{d}x = \mathrm{e}^{x} - \ln\left(\mathrm{e}^{x} + 1\right) + k\] [5 marks]
(b) Solve the differential equation\[\left(\frac{\mathrm{e}^{x} + 1}{\mathrm{e}^{2x}}\right)\frac{\mathrm{d}y}{\mathrm{d}x} = \cos^2 y\]given that \(y = \pi\) when \(x = 0\)
Write your answer in the form
\[\tan y = \mathrm{e}^{x} + \ln\left(\frac{A}{\mathrm{e}^{x} + 1}\right) + B\]where \(A\) and \(B\) are constants to be found.
[5 marks]| Scheme | Marks | AO |
|---|---|---|
| Obtains \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{e}^{x}\) OE | B1 | 1.1b |
| Makes a partial substitution to obtain \(\displaystyle\int \frac{\mathrm{e}^{2x}}{u}\frac{1}{\mathrm{e}^{x}}\,\mathrm{d}u\) or better If they simplify the fraction first \(\displaystyle\int \mathrm{e}^{x} - \frac{\mathrm{e}^{x}}{\mathrm{e}^{x} + 1}\,\mathrm{d}x\) need to see \(\displaystyle\int \mathrm{e}^{x} - \frac{\mathrm{e}^{x}}{u}\frac{1}{\mathrm{e}^{x}}\,\mathrm{d}u\) or better for this mark | M1 | 3.1a |
| Obtains \(\displaystyle\int \frac{u - 1}{u}\,\mathrm{d}u\) OE | A1 | 1.1b |
| Integrates their two-term integrand, which is in terms of u, with at least one of their terms integrated correctly. | M1 | 1.1a |
| Completes reasoned argument to obtain \(\mathrm{e}^{x} - \ln\left(\mathrm{e}^{x} + 1\right) + k\) Must see constant of integration introduced as they integrate and dealt with consistently AG | R1 | 2.1 |
| (5) |
Typical solution
\[u = \mathrm{e}^{x} + 1\]\[\frac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{e}^{x}\]\[\int \frac{\mathrm{e}^{2x}}{\mathrm{e}^{x} + 1}\,\mathrm{d}x = \int \frac{\mathrm{e}^{2x}}{u}\frac{1}{\mathrm{e}^{x}}\,\mathrm{d}u\]\[= \int \frac{\mathrm{e}^{x}}{u}\,\mathrm{d}u\]\[= \int \frac{u - 1}{u}\,\mathrm{d}u\]\[= \int 1 - \frac{1}{u}\,\mathrm{d}u\]\[= u - \ln u + c\]\[= \mathrm{e}^{x} + 1 - \ln\left(\mathrm{e}^{x} + 1\right) + c\]\[= \mathrm{e}^{x} - \ln\left(\mathrm{e}^{x} + 1\right) + k\]| Scheme | Marks | AO |
|---|---|---|
| Obtains \(\dfrac{1}{\cos^2 y}\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{e}^{2x}}{\mathrm{e}^{x} + 1}\) or better | B1 | 3.1a |
| Integrates to obtain one correct side. Uses given answer from (a) or recalls \(\displaystyle\int \sec^2 y\,\mathrm{d}y = \tan y\) | M1 | 1.1a |
| Obtains \(\tan y = \mathrm{e}^{x} - \ln\left(\mathrm{e}^{x} + 1\right) + k\) Condone missing or extra constants of integration. | A1 | 1.1b |
| Substitutes \(y = \pi\) and \(x = 0\) into their integrated equation which must be formed from exponential functions of \(x\) and trigonometric function(s) of \(y\) to obtain the constant of integration | M1 | 1.1a |
| Completes reasoned argument to obtain \(\tan y = \mathrm{e}^{x} + \ln\left(\dfrac{2}{\mathrm{e}^{x} + 1}\right) - 1\) | R1 | 2.1 |
| (5) | ||
| (10 marks) |