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3. (a) Find \( \int x \cos 2x \, dx \) - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 8
Question 6
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 6 - 2007 - Paper 8
Step 1
Find \( \int x \cos 2x \, dx \)
Answer
To solve ( \int x \cos 2x , dx ), we will use integration by parts, where we let:
- ( u = x ) and ( dv = \cos 2x , dx )
Then, differentiate and integrate accordingly:
- ( du = dx )
- ( v = \frac{1}{2} \sin 2x ) (since ( \int \cos 2x , dx = \frac{1}{2} \sin 2x + C ))
Now using the integration by parts formula, ( \int u , dv = uv - \int v , du ), we have:
[ \int x \cos 2x , dx = x \cdot \frac{1}{2} \sin 2x - \int \frac{1}{2} \sin 2x , dx ]
Next, we integrate ( \int \sin 2x , dx ):
- ( \int \sin 2x , dx = -\frac{1}{2} \cos 2x + C )
So, substituting back:
[ \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
To find ( \int x \cos^3 x , dx ), we can apply the identity for ( \cos 2x ):
[ \cos 2x = 2 \cos^2 x - 1 \Rightarrow \cos^2 x = \frac{1 + \cos 2x}{2} ]
Thus: [ \cos^3 x = \cos x \cdot \cos^2 x = \cos x \cdot \frac{1 + \cos 2x}{2} ]
Now, expanding the integral gives: [ \int x \cos^3 x , dx = \frac{1}{2} \int x \cos x , dx + \frac{1}{2} \int x \cos 2x , dx ]
The second integral has already been solved in part (a), while the first can also involve integration by parts, eventually leading to: [ \int x \cos^3 x , dx = \frac{1}{2} \left( \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x \right) + C ]