Edexcel IAL October 2022 P4 (WMA14/01) Q7 Integration Substitution
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
(i) Use the substitution \( u = e^x – 3 \) to show that
\[\int_{\ln5}^{\ln7} \frac{4e^{3x}}{e^x – 3} dx = a + b \ln 2\]
where a and b are constants to be found.
(7)
(ii) Show, by integration, that
\[\int 3e^x \cos 2x dx = pe^x \sin 2x + qe^x \cos 2x + c\]
where p and q are constants to be found and c is an arbitrary constant.
(5)
Solution to Part i: Integration by Substitution
Key Concepts Used:
-
Substitution Method: Used to simplify the integrand by introducing a new variable \( u \).
- Change of Limits: Adjusting the integration limits to match the new variable.
- Logarithmic Integration: Integrating terms of the form \(\frac{1}{u}\).
Step-by-Step Solution:
1. Substitution Setup
Given the substitution:
\[u = e^x – 3\]
Express \( e^x \) in terms of \( u \):
\[e^x = u + 3\]
2. Compute \( du \) and \( dx \)
Differentiate both sides with respect to \( x \):
3. Rewrite the Integral in Terms of \( u \)
Original integral:
\[\int_{\ln5}^{\ln7} \frac{4e^{3x}}{e^x – 3} dx\]
Substitute \( e^x = u + 3 \) and \( dx = \frac{du}{u+3} \):
\[\int \frac{4(u + 3)^3}{u} \cdot \frac{du}{u + 3} = \int \frac{4(u + 3)^2}{u} du\]
4. Expand the Integrand
Expand \((u + 3)^2\):
\[(u + 3)^2 = u^2 + 6u + 9\]
\[\frac{4(u^2 + 6u + 9)}{u} = 4u + 24 + \frac{36}{u}\]
5. Integrate Term by Term
\[\int \left( 4u + 24 + \frac{36}{u} \right) du = 2u^2 + 24u + 36\ln|u| + C\]
6. Change the Limits of Integration
Original limits:
When \( x = \ln 5 \):
\[u = 5 – 3 = 2\]
When \( x = \ln 7 \):
\[u = 7 – 3 = 4\]
7. Evaluate the Definite Integral
\[\left[2u^2 + 24u + 36\ln u\right]_{2}^{4}\]
At \( u = 4 \):
\[32 + 96 + 36\ln 4\]
At \( u = 2 \):
\[8 + 48 + 36\ln 2\]
Subtract:
\[(128 + 36\ln 4) – (56 + 36\ln 2) = 72 + 36(\ln 4 – \ln 2)\]
Simplify:
\[\ln 4 – \ln 2 = \ln 2\]
\[72 + 36\ln 2\]
Final Answer:
\[72 + 36\ln 2\]
Solution to Part ii: Integration by Parts
Key Concepts Used:
- Integration by Parts: Used to integrate products of functions, given by \(\int u dv = uv – \int v du\).
- Cyclic Integration: Reapplying integration by parts to resolve recurring integrals.
- Solving for the Integral: Combining like terms to isolate the desired integral.
Step-by-Step Solution:
1. First Integration by Parts
Let:
\[u = \cos 2x \\
dv = 3e^x dx\]
Then:
1. First Integration by Parts
Let:
\[u = \cos 2x \\
dv = 3e^x dx\]
Then:
\[du = -2\sin 2x dx \\
v = 3e^x\]
Apply integration by parts:
\[\int 3e^x \cos 2x dx = 3e^x \cos 2x – \int 3e^x (-2\sin 2x) dx = 3e^x \cos 2x + 6 \int e^x \sin 2x dx\]
2. Second Integration by Parts
Now, let:
\[u = \sin 2x \\
dv = e^x dx\]
Then:
\[du = 2\cos 2x dx \\
v = e^x\]
Apply integration by parts again:
\[\int e^x \sin 2x dx = e^x \sin 2x – \int e^x (2\cos 2x) dx\]
\[= e^x \sin 2x – 2 \int e^x \cos 2x dx\]
3. Substitute Back
From Step 1:
\[\int 3e^x \cos 2x dx = 3e^x \cos 2x + 6 \left( e^x \sin 2x – 2 \int e^x \cos 2x dx \right)\]
\[= 3e^x \cos 2x + 6e^x \sin 2x – 12 \int e^x \cos 2x dx\]
Let \( I = \int 3e^x \cos 2x dx \):
\[I = 3e^x \cos 2x + 6e^x \sin 2x – 4I\]
4. Solve for \( I \)
Combine like terms:
\[5I = 3e^x \cos 2x + 6e^x \sin 2x\]
\[I = \frac{3}{5} e^x \cos 2x + \frac{6}{5} e^x \sin 2x + C\]
Final Answer:
\[\frac{6}{5} e^x \sin 2x + \frac{3}{5} e^x \cos 2x + C\]
