6. Use integration by parts to show that
$$\int \mathrm { e } ^ { 2 x } \cos 3 x \mathrm {~d} x = p \mathrm { e } ^ { 2 x } \sin 3 x + q \mathrm { e } ^ { 2 x } \cos 3 x + k$$
where \(p\) and \(q\) are rational numbers to be found and \(k\) is an arbitrary constant.
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Question 6 (Way 1):
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(\int e^{2x}\cos3x\,dx=\frac{1}{3}e^{2x}\sin3x-\frac{2}{3}\int e^{2x}\sin3x\,dx\ (+c)\) M1A1
M1: applying parts to obtain \(\alpha e^{2x}\sin3x\pm\beta\int e^{2x}\sin3x\,dx\); A1: correct expression
Applies parts again to \(\int e^{2x}\sin3x\,dx\) obtaining \(\alpha e^{2x}\cos3x\pm\beta\int e^{2x}\cos3x\,dx\) M1
\(\int e^{2x}\cos3x\,dx=\frac{1}{3}e^{2x}\sin3x+\frac{2}{9}e^{2x}\cos3x-\frac{4}{9}\int e^{2x}\cos3x\,dx\ (+c)\) A1
Fully correct application of parts twice
\(\frac{13}{9}\int e^{2x}\cos3x\,dx=\frac{1}{3}e^{2x}\sin3x+\frac{2}{9}e^{2x}\cos3x\ (+c)\) M1
Fully correct strategy for finding \(\int e^{2x}\cos3x\,dx\)
\(=\dfrac{3}{13}e^{2x}\sin3x+\dfrac{2}{13}e^{2x}\cos3x+k\) A1
cao
Question 6 (Way 2):
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(\int e^{2x}\cos3x\,dx=\frac{1}{2}e^{2x}\cos3x+\frac{3}{2}\int e^{2x}\sin3x\,dx\ (+c)\) M1A1
M1: applying parts to obtain \(\alpha e^{2x}\cos3x\pm\beta\int e^{2x}\sin3x\,dx\); A1: correct expression
Applies parts again to \(\int e^{2x}\sin3x\,dx\) obtaining \(\alpha e^{2x}\sin3x\pm\beta\int e^{2x}\cos3x\,dx\) M1
\(\int e^{2x}\cos3x\,dx=\frac{1}{2}e^{2x}\cos3x+\frac{3}{4}e^{2x}\sin3x-\frac{9}{4}\int e^{2x}\cos3x\,dx\ (+c)\) A1
Fully correct application of parts twice
\(\frac{13}{4}\int e^{2x}\cos3x\,dx=\frac{1}{2}e^{2x}\cos3x+\frac{3}{4}e^{2x}\sin3x\ (+c)\) M1
Fully correct strategy for finding \(\int e^{2x}\cos3x\,dx\)
\(=\dfrac{3}{13}e^{2x}\sin3x+\dfrac{2}{13}e^{2x}\cos3x+k\) A1
cao
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## Question 6 (Way 1):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int e^{2x}\cos3x\,dx=\frac{1}{3}e^{2x}\sin3x-\frac{2}{3}\int e^{2x}\sin3x\,dx\ (+c)$ | M1A1 | M1: applying parts to obtain $\alpha e^{2x}\sin3x\pm\beta\int e^{2x}\sin3x\,dx$; A1: correct expression |
| Applies parts again to $\int e^{2x}\sin3x\,dx$ obtaining $\alpha e^{2x}\cos3x\pm\beta\int e^{2x}\cos3x\,dx$ | M1 | |
| $\int e^{2x}\cos3x\,dx=\frac{1}{3}e^{2x}\sin3x+\frac{2}{9}e^{2x}\cos3x-\frac{4}{9}\int e^{2x}\cos3x\,dx\ (+c)$ | A1 | Fully correct application of parts twice |
| $\frac{13}{9}\int e^{2x}\cos3x\,dx=\frac{1}{3}e^{2x}\sin3x+\frac{2}{9}e^{2x}\cos3x\ (+c)$ | M1 | Fully correct strategy for finding $\int e^{2x}\cos3x\,dx$ |
| $=\dfrac{3}{13}e^{2x}\sin3x+\dfrac{2}{13}e^{2x}\cos3x+k$ | A1 | cao |
## Question 6 (Way 2):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int e^{2x}\cos3x\,dx=\frac{1}{2}e^{2x}\cos3x+\frac{3}{2}\int e^{2x}\sin3x\,dx\ (+c)$ | M1A1 | M1: applying parts to obtain $\alpha e^{2x}\cos3x\pm\beta\int e^{2x}\sin3x\,dx$; A1: correct expression |
| Applies parts again to $\int e^{2x}\sin3x\,dx$ obtaining $\alpha e^{2x}\sin3x\pm\beta\int e^{2x}\cos3x\,dx$ | M1 | |
| $\int e^{2x}\cos3x\,dx=\frac{1}{2}e^{2x}\cos3x+\frac{3}{4}e^{2x}\sin3x-\frac{9}{4}\int e^{2x}\cos3x\,dx\ (+c)$ | A1 | Fully correct application of parts twice |
| $\frac{13}{4}\int e^{2x}\cos3x\,dx=\frac{1}{2}e^{2x}\cos3x+\frac{3}{4}e^{2x}\sin3x\ (+c)$ | M1 | Fully correct strategy for finding $\int e^{2x}\cos3x\,dx$ |
| $=\dfrac{3}{13}e^{2x}\sin3x+\dfrac{2}{13}e^{2x}\cos3x+k$ | A1 | cao |
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6. Use integration by parts to show that
$$\int \mathrm { e } ^ { 2 x } \cos 3 x \mathrm {~d} x = p \mathrm { e } ^ { 2 x } \sin 3 x + q \mathrm { e } ^ { 2 x } \cos 3 x + k$$
where $p$ and $q$ are rational numbers to be found and $k$ is an arbitrary constant.\\
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\section*{Question 6 continued}
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\hfill \mbox{\textit{Edexcel P4 2022 Q6 [6]}}