Integration by Substitution

Edexcel IAL January 2021 P4 (WMA14/01) Q5 Integration by Substitution

In this question you should show all stages of your working.
Solutions relying on calculator technology are not acceptable.

Using the substitution \( u = 3 + \sqrt{2x – 1} \) find the exact value of

\[\int_{1}^{13} \frac{4}{3 + \sqrt{2x – 1}} dx\]

giving your answer in the form p + q ln 2, where p and q are integers to be found.

Solution: Setting Up the Integral and Evaluating the Integral

Key Concepts Used:

Algebraic Substitution: Using \( u = 3 + \sqrt{2x – 1} \) to simplify the integrand.
Implicit Differentiation: Finding the relationship between \( du \) and \( dx \).
Limit Transformation: Adjusting integration limits to match the new variable.
Basic Integration Rules:
- \( \int 4du = 4u \)
- \( \int \frac{12}{u} du = 12 \ln|u| \)
Logarithmic Properties: Simplifying logarithmic expressions.
Exact Value Calculation: Evaluating definite integrals.

Step-by-Step Solution:

1. Apply the Given Substitution

Given substitution:

\[u = 3 + \sqrt{2x – 1}\]

Express the square root term:

\[u – 3 = \sqrt{2x – 1}\]

2. Compute the Differential

Square both sides:

\[(u – 3)^2 = 2x – 1\]

Differentiate implicitly with respect to \( x \):

\[2(u – 3)\frac{du}{dx} = 2\]

Cancel 2 on both sides of equation:

From previous step:

\[\frac{du}{dx} = \frac{1}{u-3}\]

\[dx = (u-3)du\]

3. Rewrite the Original Integral

Original integral:

\[\int_{1}^{13} \frac{4}{3 + \sqrt{2x – 1}} dx\]

Substitute \( u \) and \( dx \):

\[\int \frac{4}{u} \cdot (u-3) du = \int \left( 4 – \frac{12}{u} \right) du\]

4. Change the Integration Limits

Original limits \( x = 1 \) to \( x = 13 \):

When \( x = 1 \): \( u = 3 + \sqrt{2(1) – 1} = 4 \)

When \( x = 13 \): \( u = 3 + \sqrt{2(13) – 1} = 8 \)

The transformed integral is:

\[\int_{4}^{8} \left( 4 – \frac{12}{u} \right) du\]

5. Integrate the Simplified Expression

\[\int \left( 4 – \frac{12}{u} \right) du = 4u – 12 \ln |u| + C\]

6. Evaluate at the New Limits

At \( u = 8 \): \( 4(8) – 12 \ln 8 = 32 – 12 \ln 8 \)

At \( u = 4 \): \( 4(4) – 12 \ln 4 = 16 – 12 \ln 4 \)

7. Compute the Definite Integral

Subtract lower limit from upper limit:

\[(32 – 12 \ln 8) – (16 – 12 \ln 4) = 16 – 12 (\ln 8 – \ln 4)\]

8. Simplify the Logarithmic Term

Using logarithm properties:

\[\ln 8 – \ln 4 = \ln \left( \frac{8}{4} \right) = \ln 2\]

Thus:

\[16 – 12 \ln 2\]

Final Answer:

The exact value of the integral is: \[16 – 12 \ln 2\]

Coach Name: Sir Muhammad Abdullah Shah

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