18M.1.AHL.TZ2.H_8

pestleMathematicsAAHLPaper 118M· ahl-5-16-integration-by-substitution-parts-and-repeated-partssource ↗

Use the substitution $u = {x^{\frac{1}{2}}}$ to find $\int {\frac{{{\text{d}}x}}{{{x^{\frac{3}{2}}} + {x^{\frac{1}{2}}}}}}$ .

[4]

Hence find the value of $\frac{1}{2}\int\limits_1^9 {\frac{{{\text{d}}x}}{{{x^{\frac{3}{2}}} + {x^{\frac{1}{2}}}}}}$ , expressing your answer in the form arctan $q$ , where $q \in \mathbb{Q}$ .

[3]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{2}{x^{ - \frac{1}{2}}}$ (accept ${\text{d}}u = \frac{1}{2}{x^{ - \frac{1}{2}}}{\text{d}}x$ or equivalent) A1

substitution, leading to an integrand in terms of $u$ M1

$\int {\frac{{2u{\text{d}}u}}{{{u^3} + u}}}$ or equivalent A1

= 2 arctan $\left( {\sqrt x } \right)\left( { + c} \right)$ A1

[4 marks]

$\frac{1}{2}\int\limits_1^9 {\frac{{{\text{d}}x}}{{{x^{\frac{3}{2}}} + {x^{\frac{1}{2}}}}}}$ = arctan 3 − arctan 1 A1

tan(arctan 3 − arctan 1) = $\frac{{3 - 1}}{{1 + 3 \times 1}}$ (M1)

tan(arctan 3 − arctan 1) = $\frac{1}{2}$

arctan 3 − arctan 1 = arctan $\frac{1}{2}$ A1

[3 marks]

Examiners’ report

[N/A]