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2. (a) Use integration by parts to find \( \int x \sin 3x \, dx \) - Edexcel - A-Level Maths Pure - Question 4 - 2012 - Paper 8
Question 4
2. (a) Use integration by parts to find \( \int x \sin 3x \, dx \). (b) Using your answer to part (a), find \( \int x \cos 3x \, dx \).
Worked Solution & Example Answer:2. (a) Use integration by parts to find \( \int x \sin 3x \, dx \) - Edexcel - A-Level Maths Pure - Question 4 - 2012 - Paper 8
Step 1
Use integration by parts to find \( \int x \sin 3x \, dx \)
Answer
To solve ( \int x \sin 3x , dx ) using integration by parts, we apply the formula:
[ \int u , dv = uv - \int v , du ]
Let:
- ( u = x ) → ( du = dx )
- ( dv = \sin 3x , dx ) → ( v = -\frac{1}{3} \cos 3x )
Plugging these into the integration by parts formula, we have:
[ \int x \sin 3x , dx = -\frac{1}{3} x \cos 3x - \int -\frac{1}{3} \cos 3x , dx ]
Now, we evaluate the remaining integral:
[ \int \cos 3x , dx = \frac{1}{3} \sin 3x + C ]
Thus, substituting this back, we get:
[ \int x \sin 3x , dx = -\frac{1}{3} x \cos 3x + \frac{1}{9} \sin 3x + C ]
Step 2
Using your answer to part (a), find \( \int x \cos 3x \, dx \)
Answer
Using the result from part (a), we can find ( \int x \cos 3x , dx ) again using integration by parts:
Let:
- ( u = x ) → ( du = dx )
- ( dv = \cos 3x , dx ) → ( v = \frac{1}{3} \sin 3x )
Now apply the formula:
[ \int x \cos 3x , dx = \frac{1}{3} x \sin 3x - \int \frac{1}{3} \sin 3x , dx ]
From part (a), we know: [ \int \sin 3x , dx = -\frac{1}{3} \cos 3x + C ]
Thus substituting back, we have:
[ \int x \cos 3x , dx = \frac{1}{3} x \sin 3x + \frac{1}{9} \cos 3x + C ]