IBDP Maths AHL 5.16 Integration by substitution AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Let \( u = \ln y \Rightarrow \text{d}u = \frac{1}{y} \text{d}y \) (A1)(A1).
\[\int \frac{\tan (\ln y)}{y} \text{d}y = \int \tan u \, \text{d}u \] (A1).
\[ = \int \frac{\sin u}{\cos u} \text{d}u = -\ln |\cos u| + c \] (A1).
EITHER
\[\int \frac{\tan (\ln y)}{y} \text{d}y = -\ln |\cos (\ln y)| + c\] (A1)(A1).
OR
\[\int \frac{\tan (\ln y)}{y} \text{d}y = \ln |\sec (\ln y)| + c\] (A1)(A1).
[6 marks]
▶️ Answer/Explanation
(a) 
Correct shape and domain (A1). Local maximum at \( (1.41, 0.0884) \) or \( \left( \sqrt{2}, \frac{\sqrt{2}}{16} \right) \) (A1). [2 marks]
(b) EITHER \( u = t^2 \), \( \frac{\text{d}u}{\text{d}t} = 2t \) (A1) OR \( t = u^{\frac{1}{2}} \), \( \frac{\text{d}t}{\text{d}u} = \frac{1}{2} u^{-\frac{1}{2}} \) (A1). \[ \int \frac{t}{12 + t^4} \text{d}t = \frac{1}{2} \int \frac{\text{d}u}{12 + u^2} \] (M1). \[ = \frac{1}{2 \sqrt{12}} \arctan \left( \frac{u}{\sqrt{12}} \right) (+ c) \] (M1). \[ = \frac{1}{4 \sqrt{3}} \arctan \left( \frac{t^2}{2 \sqrt{3}} \right) (+ c) \] (A1). [4 marks]
(c) \[ \int_0^6 \frac{t}{12 + t^4} \text{d}t \] (M1). \[ = \left[ \frac{1}{4 \sqrt{3}} \arctan \left( \frac{t^2}{2 \sqrt{3}} \right) \right]_0^6 \] (M1). \[ = \frac{1}{4 \sqrt{3}} \arctan \left( \frac{18}{\sqrt{3}} \right) \] or \( \frac{\sqrt{3}}{12} \arctan (6 \sqrt{3}) \) (A1). [3 marks]
(d) \[ \frac{\text{d}v}{\text{d}s} = \frac{1}{2 \sqrt{s (1 – s)}} \] (A1). \[ a = v \frac{\text{d}v}{\text{d}s} = \arcsin(\sqrt{s}) \cdot \frac{1}{2 \sqrt{s (1 – s)}} \] (M1). At \( s = 0.1 \): \[ a = \arcsin(\sqrt{0.1}) \cdot \frac{1}{2 \sqrt{0.1 \cdot 0.9}} = 0.536 \, \text{m s}^{-2} \] (A1). [3 marks]
Examiners Report:
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