Standard +0.8 This is a Further Maths FP3 question requiring knowledge of complex exponentials and integration. While the technique (combining integrals using Euler's formula) is elegant once known, it requires conceptual sophistication beyond standard A-level. The calculation itself is moderately involved with complex arithmetic and careful evaluation of limits, but follows a clear path once the method is understood. Slightly above average difficulty for Further Maths material.
The integrals \(C\) and \(S\) are defined by
$$C = \int_0^{3\pi} e^{3x} \cos 3x \, dx \quad \text{and} \quad S = \int_0^{3\pi} e^{3x} \sin 3x \, dx.$$
By considering \(C + iS\) as a single integral, show that
$$C = -\frac{1}{3}(2 + 3e^{\pi}),$$
and obtain a similar expression for \(S\). [8]
(You may assume that the standard result for \(\int e^{kx} dx\) remains true when \(k\) is a complex constant, so that \(\int e^{(a+ib)x} dx = \frac{1}{a+ib}e^{(a+ib)x}\).)
The integrals $C$ and $S$ are defined by
$$C = \int_0^{3\pi} e^{3x} \cos 3x \, dx \quad \text{and} \quad S = \int_0^{3\pi} e^{3x} \sin 3x \, dx.$$
By considering $C + iS$ as a single integral, show that
$$C = -\frac{1}{3}(2 + 3e^{\pi}),$$
and obtain a similar expression for $S$. [8]
(You may assume that the standard result for $\int e^{kx} dx$ remains true when $k$ is a complex constant, so that $\int e^{(a+ib)x} dx = \frac{1}{a+ib}e^{(a+ib)x}$.)
\hfill \mbox{\textit{OCR FP3 2008 Q4 [8]}}