Edexcel C4 — Question 2 6 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeDefinite integral with simple linear/polynomial substitution
DifficultyModerate -0.3 This is a straightforward substitution question where the substitution is given explicitly and the integrand is already in the perfect form for it (cos x is the derivative of sin x). Students need to change limits, integrate u³, and substitute back—all standard C4 techniques with no problem-solving required, making it slightly easier than average.
Spec1.08h Integration by substitution
Use the substitution \(u = 1 + \sin x\) to find the value of $$\int_0^{\frac{\pi}{4}} \cos x (1 + \sin x)^3 \, dx.$$ [6]
AnswerMarks Guidance
\(u = 1 + \sin x \Rightarrow \frac{du}{dx} = \cos x\)M1
\(x = 0 \Rightarrow u = 1, x = \frac{\pi}{2} \Rightarrow u = 2\)B1
\(I = \int_1^2 u^3 \, du\)A1
\(= \left[\frac{1}{4}u^4\right]_1^2\)M1
\(= 4 - \frac{1}{4} = \frac{15}{4}\)M1 A1 (6) 
$u = 1 + \sin x \Rightarrow \frac{du}{dx} = \cos x$ | M1 |
$x = 0 \Rightarrow u = 1, x = \frac{\pi}{2} \Rightarrow u = 2$ | B1 |
$I = \int_1^2 u^3 \, du$ | A1 |
$= \left[\frac{1}{4}u^4\right]_1^2$ | M1 |
$= 4 - \frac{1}{4} = \frac{15}{4}$ | M1 A1 | (6)
Use the substitution $u = 1 + \sin x$ to find the value of
$$\int_0^{\frac{\pi}{4}} \cos x (1 + \sin x)^3 \, dx.$$ [6]

\hfill \mbox{\textit{Edexcel C4  Q2 [6]}}
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