| Exam Board | SPS |
|---|---|
| Module | SPS FM Pure (SPS FM Pure) |
| Year | 2025 |
| Session | June |
| Marks | 9 |
| Topic | Integration by Substitution |
| Type | Trigonometric substitution: direct evaluation |
| Difficulty | Challenging +1.8 Part (a) is a standard Further Maths integral requiring the substitution x = sin θ, which is routine for FM students. Part (b) is significantly harder: the RHS factorizes as 2√[(1-x²)(1-y²)], suggesting separation of variables, but verifying the given solution requires computing a complex derivative involving products and chains of √(1-x²) terms and checking it matches the factorized form. This verification demands careful algebraic manipulation and is non-trivial, though the factorization hint makes it more accessible than a pure discovery problem. |
| Spec | 1.08h Integration by substitution1.08k Separable differential equations: dy/dx = f(x)g(y) |
\begin{enumerate}[label=(\alph*)]
\item Using a suitable substitution, find
$$\int \sqrt{1 - x^2} \, dx.$$
[4]
\item Show that the differential equation
$$\frac{dy}{dx} = 2\sqrt{1 - x^2 - y^2 + x^2y^2},$$
given that $y = 0$ when $x = 0$, $|x| < 1$ and $|y| < 1$, has the solution
$$y = x \cos\left(x\sqrt{1 - x^2}\right) + \sqrt{1 - x^2} \sin\left(x\sqrt{1 - x^2}\right).$$
[5]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM Pure 2025 Q13 [9]}}