Edexcel IAL June 2022 P4 (WMA14/01) Q5 Integration Substitution
In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.
(a) Use the substitution \( x = 2 \sin u \) to show that
\[\int_{0}^{1} \frac{3x + 2}{(4 – x^2)^{\frac{3}{2}}} dx = \int_{0}^{p} \left( \frac{3}{2} \sec u \tan u + \frac{1}{2} \sec^2 u \right) du\]
where \( p \) is a constant to be found.
(b) Hence find the exact value of
\[\int_{0}^{1} \frac{3x + 2}{(4 – x^2)^{\frac{3}{2}}} dx\]
Solution to Part a: Transforming the Integral
Key Concepts Used:
- Trigonometric Substitution: Used to simplify integrals containing expressions like \((a^2 – x^2)^{n/2}\).
- Trigonometric Identities: Pythagorean identity \(\sin^2 u + \cos^2 u = 1\) is applied.
- Limit Transformation: Changing integration limits to match the new variable.
Step-by-Step Solution:
1. Apply the Substitution
Given substitution:
\[x = 2 \sin u\]
Compute differential:
\[\frac{dx}{du} = 2 \cos u\]
\[dx = 2 \cos u du\]
2. Rewrite the Integrand
Original integrand:
\[\frac{3x + 2}{(4 – x^2)^{3/2}}\]
Substitute \( x = 2 \sin u \):
Numerator:
\[3(2 \sin u) + 2 = 6 \sin u + 2\]
Denominator:
\[4 – (2 \sin u)^2 = 4 – 4 \sin^2 u\]
\[= 4 – 4(1 – \cos^2 u)\]
\[= 4 – 4 + 4\cos^2u\]
\[= 4\cos^2u\]
Thus:
\[(4\cos^2u)^{3/2} = 8\cos^3u\]
New integrand:
\[\frac{6\sin u + 2}{8\cos^3u} \cdot 2\cos u = \frac{12\sin u + 4}{8\cos^2u}\]
3. Simplify the Expression
Separate terms:
\[\frac{12\sin u}{8\cos^2u} + \frac{4}{8\cos^2u} = \frac{3\sin u}{2\cos^2u} + \frac{1}{2\cos^2u}\]
Express using secant and tangent:
\[\frac{3}{2}\sec u \tan u + \frac{1}{2}\sec^2 u\]
4. Change the Limits
Original limits \( x = 0 \) to \( x = 1 \):
When \( x = 0 \): \( 2\sin u = 0 \Rightarrow u = 0 \)
When \( x = 1 \): \( 2\sin u = 1 \Rightarrow u = \frac{\pi}{6} \)
Final Answer:
\[\int_{0}^{\frac{\pi}{6}} \left( \frac{3}{2}\sec u \tan u + \frac{1}{2}\sec^2 u \right) du\]
Solution to Part b: Evaluating the Integral
Key Concepts Used:
- Basic Trigonometric Integrals:
- \(\int \sec u \tan u du = \sec u\)
- \(\int \sec^2 u du = \tan u\)
- Exact Value Evaluation: Using known trigonometric values at specific angles.
Step-by-Step Solution:
1. Integrate the Simplified Expression
From part (a):
\[\int \left( \frac{3}{2}\sec u \tan u + \frac{1}{2}\sec^2 u \right) du\]
1. Integrate the Simplified Expression
Integrate term by term:
\[\int \frac{3}{2} \sec u \tan u du = \frac{3}{2} \sec u\]
\[\int \frac{1}{2} \sec^2 u du = \frac{1}{2} \tan u\]
\[\frac{3}{2} \sec u + \frac{1}{2} \tan u + C\]
2. Evaluate at Limits
At \( u = \frac{\pi}{6} \):
\[\sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}, \quad \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\]
\[\frac{3}{2} \sec u + \frac{1}{2} \tan u = \frac{3}{2} \cdot \frac{2}{\sqrt{3}} + \frac{1}{2} \cdot \frac{1}{\sqrt{3}}\]
\[= \frac{3}{\sqrt{3}} + \frac{1}{2\sqrt{3}} = \frac{7}{2\sqrt{3}}\]
At \( u = 0 \):
\[\sec 0 = 1, \quad \tan 0 = 0\]
\[\frac{3}{2} \sec u + \frac{1}{2} \tan u = \frac{3}{2} \cdot 1 + \frac{1}{2} \cdot 0 = \frac{3}{2}\]
3. Compute the Definite Integral
Subtract lower limit from upper limit:
\[\frac{7}{2\sqrt{3}} – \frac{3}{2}\]
Rationalize:
\[\frac{7}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{7\sqrt{3}}{6}\]
\[\frac{7}{2\sqrt{3}} – \frac{3}{2} = \frac{7\sqrt{3}}{6} – \frac{3}{2}\]
\[= \frac{7\sqrt{3} – 9}{6}\]
Final Answer:
\[\frac{7\sqrt{3} – 9}{6}\]