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3. (a) Find \( \int x \cos{2x} \, dx \) - Edexcel - A-Level Maths Pure - Question 5 - 2007 - Paper 7
Question 5
3. (a) Find \( \int x \cos{2x} \, dx \). (b) Hence, using the identity \( \cos{2x} = 2\cos^{2}x - 1 \), deduce \( \int x \cos^{2}x \, dx \).
Worked Solution & Example Answer:3. (a) Find \( \int x \cos{2x} \, dx \) - Edexcel - A-Level Maths Pure - Question 5 - 2007 - Paper 7
Step 1
Find \( \int x \cos{2x} \, dx \)
Answer
To solve ( \int x \cos{2x} , dx ), we use integration by parts. Let:
[ u = x , \Rightarrow , du = dx \ [ dv = \cos{2x} , dx \rightarrow v = \frac{1}{2}\sin{2x} ]
Using the integration by parts formula:
[ \int u , dv = uv - \int v , du ]
we get:
[ \int x \cos{2x} , dx = x \cdot \frac{1}{2}\sin{2x} - \int \frac{1}{2}\sin{2x} , dx ]
Now, we calculate ( \int \frac{1}{2}\sin{2x} , dx ):
[ \int \frac{1}{2}\sin{2x} , dx = -\frac{1}{4}\cos{2x} + C ]
Thus, combining the results:
[ \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 \( \int x \cos^{2}x \, dx \)
Answer
To deduce ( \int x \cos^{2}x , dx ) using the identity ( \cos{2x} = 2\cos^{2}x - 1 ), we rewrite the integral:
[ \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 ( \int x \cos{2x} , dx = \frac{x}{2}\sin{2x} + \frac{1}{4}\cos{2x} + C ).
Also, the integral ( \int x , dx = \frac{x^{2}}{2} + C ).
Combining both results:
[ \int x \cos^{2}x , dx = \frac{1}{2}\left( \frac{x}{2}\sin{2x} + \frac{1}{4}\cos{2x} + C \right) + \frac{1}{2}\left( \frac{x^{2}}{2} + C \right) ]
This simplifies to:
[ = \frac{x}{4}\sin{2x} + \frac{1}{8}\cos{2x} + \frac{x^{2}}{4} + C ]