Question 7 - A-Level Maths

OCR C4 — Question 7 11 marks

Exam Board	OCR
Module	C4 (Core Mathematics 4)
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Integration using inverse trig and hyperbolic functions
Type	Standard integral of 1/√(a²-x²)
Difficulty	Standard +0.3 This is a straightforward two-part integration question testing standard C4 techniques. Part (i) is a guided substitution with a simple integrand that becomes trivial after substitution, and part (ii) is a routine integration by parts. Both are textbook exercises requiring method recall rather than problem-solving, making this slightly easier than average.
Spec	1.08h Integration by substitution 1.08i Integration by parts

Use the substitution $x = 2 \sin u$ to evaluate $$\int_0^{\sqrt{3}} \frac{1}{\sqrt{4-x^2}} \, dx.$$ [6]
Evaluate $$\int_0^{\frac{\pi}{2}} x \cos x \, dx.$$ [5]

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(i)

Answer	Marks
$x = 2 \sin u \Rightarrow \frac{dx}{du} = 2 \cos u$	M1
$x = 0 \Rightarrow u = 0$, $x = \sqrt{3} \Rightarrow u = \frac{\pi}{3}$	B1
$I = \int_0^{\pi/3} \frac{1}{2\cos u} \times 2\cos u \, du = \int_0^{\pi/3} 1 \, du$	M1 A1
$= [u]_0^{\pi/3} = \frac{\pi}{3} - 0 = \frac{\pi}{3}$	M1 A1

(ii)

Answer	Marks	Guidance
$u = x$, $u' = 1$, $v' = \cos x$, $v = \sin x$	M1
$I = [x \sin x]_0^{\pi/3} - \int_0^{\pi/3} \sin x \, dx$	A1
$= [x \sin x + \cos x]_0^{\pi/3}$	M1
$= \left(\frac{\pi}{3} + 0\right) - (0 + 1) = \frac{\pi}{3} - 1$	M1 A1	(11)

## (i)
$x = 2 \sin u \Rightarrow \frac{dx}{du} = 2 \cos u$ | M1 |
$x = 0 \Rightarrow u = 0$, $x = \sqrt{3} \Rightarrow u = \frac{\pi}{3}$ | B1 |
$I = \int_0^{\pi/3} \frac{1}{2\cos u} \times 2\cos u \, du = \int_0^{\pi/3} 1 \, du$ | M1 A1 |
$= [u]_0^{\pi/3} = \frac{\pi}{3} - 0 = \frac{\pi}{3}$ | M1 A1 |

## (ii)
$u = x$, $u' = 1$, $v' = \cos x$, $v = \sin x$ | M1 |
$I = [x \sin x]_0^{\pi/3} - \int_0^{\pi/3} \sin x \, dx$ | A1 |
$= [x \sin x + \cos x]_0^{\pi/3}$ | M1 |
$= \left(\frac{\pi}{3} + 0\right) - (0 + 1) = \frac{\pi}{3} - 1$ | M1 A1 | (11) |

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Show LaTeX source

\begin{enumerate}[label=(\roman*)]
\item Use the substitution $x = 2 \sin u$ to evaluate
$$\int_0^{\sqrt{3}} \frac{1}{\sqrt{4-x^2}} \, dx.$$ [6]
\item Evaluate
$$\int_0^{\frac{\pi}{2}} x \cos x \, dx.$$ [5]
\end{enumerate}

\hfill \mbox{\textit{OCR C4  Q7 [11]}}

This paper (8 questions)

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Q1 4 Q2 7 Q3 9 Q4 9 Q5 10 Q6 10 Q7 11 Q8 12

Answer	Marks
\(x = 2 \sin u \Rightarrow \frac{dx}{du} = 2 \cos u\)	M1
\(x = 0 \Rightarrow u = 0\), \(x = \sqrt{3} \Rightarrow u = \frac{\pi}{3}\)	B1
\(I = \int_0^{\pi/3} \frac{1}{2\cos u} \times 2\cos u \, du = \int_0^{\pi/3} 1 \, du\)	M1 A1
\(= [u]_0^{\pi/3} = \frac{\pi}{3} - 0 = \frac{\pi}{3}\)	M1 A1

Answer	Marks	Guidance
\(u = x\), \(u' = 1\), \(v' = \cos x\), \(v = \sin x\)	M1
\(I = [x \sin x]_0^{\pi/3} - \int_0^{\pi/3} \sin x \, dx\)	A1
\(= [x \sin x + \cos x]_0^{\pi/3}\)	M1
\(= \left(\frac{\pi}{3} + 0\right) - (0 + 1) = \frac{\pi}{3} - 1\)	M1 A1	(11)