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3. (a) Find \( \int x \cos 2x \, dx \) - Edexcel - A-Level Maths Pure - Question 5 - 2007 - Paper 8
Question 5
3. (a) Find \( \int x \cos 2x \, dx \). (b) Hence, using the identity \( \cos 2x = 2\cos^2 x - 1 \), deduce \( \int x \cos^3 x \, dx \).
Worked Solution & Example Answer:3. (a) Find \( \int x \cos 2x \, dx \) - Edexcel - A-Level Maths Pure - Question 5 - 2007 - Paper 8
Step 1
Find \( \int x \cos 2x \, dx \)
Answer
To find ( \int x \cos 2x , dx ), we will use integration by parts. Let:
- ( u = x )
- ( dv = \cos 2x , dx )
Then, we differentiate and integrate:
- ( du = dx )
- ( v = \frac{1}{2} \sin 2x )
Using integration by parts formula ( \int u , dv = uv - \int v , du ):
[\int x \cos 2x , dx = x \left( \frac{1}{2} \sin 2x \right) - \int \frac{1}{2} \sin 2x , dx]
Now, we compute ( \int \sin 2x , dx ):
[\int \sin 2x , dx = -\frac{1}{2} \cos 2x ]
Putting it all together:
[\int x \cos 2x , dx = \frac{x}{2} \sin 2x + \frac{1}{4} \cos 2x + C]
Step 2
Hence, using the identity \( \cos 2x = 2\cos^2 x - 1 \), deduce \( \int x \cos^3 x \, dx \)
Answer
To deduce ( \int x \cos^3 x , dx ), we begin by expressing ( \cos^3 x ):
[\cos^3 x = \cos x (\cos^2 x) = \cos x \left( \frac{1}{2}(1 + \cos 2x) \right) = \frac{1}{2} \cos x + \frac{1}{2} \cos x \cos 2x]
Thus:
[\int x \cos^3 x , dx = \frac{1}{2} \int x \cos x , dx + \frac{1}{2} \int x \cos x \cos 2x , dx]
We already have the result for ( \int x \cos 2x , dx ) from part (a). Now, we substitute this into the integral:
[= \frac{1}{2} \left( \text{result from part (a)} \right) + \frac{1}{4} \sin 2x + C]
Final expression yields:
[= \frac{1}{2} \cdot \left( \frac{x}{2} \sin 2x + \frac{1}{4} \cos 2x + C \right) + \frac{1}{4} \sin 2x + C]