Edexcel IAL October 2022 P4 (WMA14/01) Q7 Integration by Substitution, Integration by Parts
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
Solutions relying entirely on calculator technology are not acceptable.
(i) Use the substitution \( u = e^x – 3 \) to show that
\(\int_{\ln 5}^{\ln 7} \frac{4e^{3x}}{e^x – 3} dx = a + b \ln 2\)
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:
-
Substitution Setup
Given the substitution:\( u = e^x – 3 \)Express \( e^x \) in terms of \( u \):
\( e^x = u + 3 \) -
Compute \( du \) and \( dx \)
Differentiate both sides with respect to \( x \):\(\frac{du}{dx} = e^x\)
\(du = e^x dx\)
\(dx = \frac{du}{e^x} = \frac{du}{u + 3}\) -
Rewrite the Integral in Terms of \( u \)
Original integral:\(\int_{\ln 5}^{\ln 7} \frac{4e^{3x}}{e^{x}-3} dx\)Substitute \( e^x = u + 3 \) and \( dx = \frac{du}{u+3} \):
\(\int \frac{4(u + 3)^3}{u} \cdot \frac{du}{u + 3} = \int \frac{4(u + 3)^2}{u} du\) -
Expand the Integrand
Expand \((u + 3)^2\):\((u + 3)^2 = u^2 + 6u + 9\)Thus:
\(\frac{4(u^2 + 6u + 9)}{u} = 4u + 24 + \frac{36}{u}\) -
Integrate Term by Term
\(\int \left( 4u + 24 + \frac{36}{u} \right) du = 2u^2 + 24u + 36\ln|u| + C\)
-
Change the Limits of Integration
Original limits:\( x = \ln 5 \): \( u = 5 – 3 = 2 \)
\( x = \ln 7 \): \( u = 7 – 3 = 4 \) -
Evaluate the Definite Integral
\( [2u^2 + 24u + 36\ln u]_{2}^{4} \)
At \( u = 4 \):
\( 32 + 96 + 36\ln4 \)At \( u = 2 \):
\( 8 + 48 + 36\ln2 \)Subtract:
\( (128 + 36\ln4) – (56 + 36\ln2) = 72 + 36(\ln4 – \ln2) \)Simplify:
\( \ln4 – \ln2 = \ln2 \)
\( 72 + 36\ln2 \)
Final Answer:
\( 72 + 36\ln2 \)
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:
-
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:
\(\int 3e^x \cos 2x dx = 3e^x \cos 2x – \int 3e^x (-2\sin 2x) dx = 3e^x \cos 2x + 6 \int e^x \sin 2x dx\) -
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:
\(\int e^x \sin 2x dx = e^x \sin 2x – \int e^x (2\cos 2x) dx\)\(= e^x \sin 2x – 2 \int e^x \cos 2x dx\) -
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 \( I = \int 3e^x \cos 2x dx \):
\(I = 3e^x \cos 2x + 6e^x \sin 2x – 4I\) -
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:
\(\frac{6}{5} e^x \sin 2x + \frac{3}{5} e^x \cos 2x + C\)