A-Level Edexcel Maths Pure Simultaneous Equations: 3. (a) Find \( \int x \cos 2x \,

3. (a) Find $ \int x \cos 2x \, dx $ - Edexcel - A-Level Maths Pure - Question 5 - 2007 - Paper 7

Question 5

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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

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Answer

To solve ( \int x \cos 2x , dx ), we will use the method of integration by parts. Let:

This gives us:

Applying integration by parts, we have:

$\int u \, dv = uv - \int v \, du$

Thus,

$\int x \cos 2x \, dx = x \cdot \frac{1}{2} \sin 2x - \int \frac{1}{2} \sin 2x \, dx$

Now, we will evaluate the remaining integral:

$\int \sin 2x \, dx = -\frac{1}{2} \cos 2x + c$

So, substituting back, we get:

$\int x \cos 2x \, dx = \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x + c$

Step 2

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Answer

Using the identity ( \cos 2x = 2 \cos^2 x - 1 ), we can express ( \int x \cos^2 x , dx ) in terms of our previous result.

First, rewrite the integral as:

$\int x \cos^2 x \, dx = \int x \left( \frac{\cos 2x + 1}{2} \right) dx = \frac{1}{2} \int x \cos 2x \, dx + \frac{1}{2} \int x \, dx$

We know from part (a):

$\int x \cos 2x \, dx = \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x + c$

Now, substituting this back, we find:

$\int x \cos^2 x \, dx = \frac{1}{2} \left( \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x \right) + \frac{1}{4} x^2 + c$

Thus,

$= \frac{1}{4} x \sin 2x + \frac{1}{8} \cos 2x + \frac{1}{4} x^2 + c$