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Find \( \int x \cos 2x \, dx \) - Edexcel - A-Level Maths Pure - Question 4 - 2007 - Paper 8
Question 4
Find \( \int x \cos 2x \, dx \). Hence, using the identity \( \cos 2x = 2 \cos^2 x - 1 \), deduce \( \int x \cos^3 x \, dx \.
Worked Solution & Example Answer:Find \( \int x \cos 2x \, dx \) - Edexcel - A-Level Maths Pure - Question 4 - 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, which is expressed as:
[ \int u , dv = uv - \int v , du ]
Here, let:
- ( u = x ) and hence ( du = dx )
- ( dv = \cos 2x , dx ) and consequently ( v = \frac{1}{2} \sin 2x )
Substituting these into the integration by parts formula:
[ \int x \cos 2x , dx = x \cdot \frac{1}{2} \sin 2x - \int \frac{1}{2} \sin 2x , dx ]
Calculating the integral:
[ \int \sin 2x , dx = -\frac{1}{2} \cos 2x + C ]
Thus, we have:
[ \int x \cos 2x , dx = \frac{1}{2} x \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
Using the identity ( \cos 2x = 2 \cos^2 x - 1 ), we have:
[ \int x \cos^3 x , dx = \int x \cos x \cdot \cos^2 x , dx = \int x \cos x \cdot \frac{1 + \cos 2x}{2} , dx ]
Thus:
[ = \frac{1}{2} \int x \cos x , dx + \frac{1}{2} \int x \cos x \cdot \cos 2x , dx ]
We already found ( \int x \cos 2x , dx ) in part (a). Hence:
[ = \frac{1}{2} \left( \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x + C \right) ]
The deduction yields:
[ \int x \cos^3 x , dx = \frac{1}{4} x \sin 2x + \frac{1}{8} \cos 2x + C ]