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3. (a) Find \( \int x \cos 2x \, dx \) - Edexcel - A-Level Maths Pure - Question 4 - 2007 - Paper 7
Question 4
3. (a) Find \( \int x \cos 2x \, dx \). (b) Hence, using the identity \( \cos 2x = 2 \cos^2 x - 1 \), deduce \( \int x \cos^2 x \, dx \).
Worked Solution & Example Answer:3. (a) Find \( \int x \cos 2x \, dx \) - Edexcel - A-Level Maths Pure - Question 4 - 2007 - Paper 7
Step 1
Find \( \int x \cos 2x \, dx \).
Answer
To solve the integral ( \int x \cos 2x , dx ), we use integration by parts, where we let:
- ( u = x )
- ( dv = \cos 2x , dx )
This gives us:
- ( du = dx )
- ( v = \frac{1}{2} \sin 2x )
Applying the integration by parts formula ( \int u , dv = uv - \int v , du ):
[ \int x \cos 2x , dx = \left( x \cdot \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 + C ]
Thus, substituting back, 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 \( \int x \cos^2 x \, dx \).
Answer
Using the identity ( \cos 2x = 2 \cos^2 x - 1 ), we can express the integral as follows:
[ \int x \cos^2 x , dx = \int x \left( \frac{1}{2} (\cos 2x + 1) \right) dx = \frac{1}{2} \int x \cos 2x , dx + \frac{1}{2} \int x , dx ]
From part (a), we already found that:
[ \int x \cos 2x , dx = \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x + C ]
Also, ( \int x , dx = \frac{1}{2} x^2 + C ). Therefore:
[ \int x \cos^2 x , dx = \frac{1}{4} x \sin 2x + \frac{1}{8} \cos 2x + \frac{1}{4} \left( \frac{1}{2} x^2 \right) + C ]
This yields:
[ \int x \cos^2 x , dx = \frac{1}{4} x \sin 2x + \frac{1}{8} \cos 2x + \frac{1}{8} x^2 + C ]