Standard +0.3 This is a straightforward integration by substitution question requiring students to apply the given substitution u = cos x, rewrite sin³x in terms of u using sin²x = 1 - cos²x, change the limits, and integrate a polynomial. While it requires multiple steps and careful algebraic manipulation, the substitution is provided and the technique is standard C4 material with no novel insight needed, making it slightly easier than average.
Award also for \(\int(1-u^2)u^2\,du = \frac{1}{3}u^3 - \frac{1}{5}u^5\)
Use new limits if \(f(u)\) or original limits if resubstitution
M1
No accuracy
\(\frac{47}{480}\); AE Fraction
A1
ISW www. If A0, answer of \(0.0979\ldots \to\) M1
[6]
## Question 5:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to connect $du$ and $dx$ or find $\frac{du}{dx}$ | M1 | No accuracy; not $du = dx$ |
| $du = -\sin x\,dx$ or $\frac{du}{dx} = -\sin x$ | A1 | |
| Indefinite integral becomes $-\int(1-u^2)u^2\,(du)$ | A1 FT | FT only from $\frac{du}{dx} = \sin x$ |
| $-\int(1-u^2)u^2\,(du) = -\frac{1}{3}u^3 + \frac{1}{5}u^5$ | B1 | Award also for $\int(1-u^2)u^2\,du = \frac{1}{3}u^3 - \frac{1}{5}u^5$ |
| Use new limits if $f(u)$ or original limits if resubstitution | M1 | No accuracy |
| $\frac{47}{480}$; AE Fraction | A1 | ISW www. If A0, answer of $0.0979\ldots \to$ M1 |
| **[6]** | | |
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