A-Level Edexcel Maths Pure Applications of Differentiation: 4. (i) Find \( \int \ln(\xi) \,

4. (i) Find \( \int \ln(\xi) \, d\xi - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 8

Question 5

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4. (i) Find $ \int \ln(\xi) \, d\xi. $ (ii) Find the exact value of $ \int_{0}^{\frac{\pi}{2}} \sin^{2} x \, dx. $

Worked Solution & Example Answer:4. (i) Find \( \int \ln(\xi) \, d\xi - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 8

Step 1

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Answer

To solve the integral ( \int \ln(\xi) , d\xi ), we will use integration by parts. Let:

Applying integration by parts:

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

Substituting in the values:

$\int \ln(\xi) \, d\xi = \xi \ln(\xi) - \int \xi \cdot \frac{1}{\xi} \, d\xi$

This simplifies to:

$= \xi \ln(\xi) - \int 1 \, d\xi = \xi \ln(\xi) - \xi + C$

Thus, the result is:

$\int \ln(\xi) \, d\xi = \xi \ln(\xi) - \xi + C$

Step 2

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Answer

To find the integral ( \int_{0}^{\frac{\pi}{2}} \sin^{2} x , dx ), we use the identity:

$\sin^{2} x = \frac{1 - \cos(2x)}{2}$

Thus, we rewrite the integral:

$\int_{0}^{\frac{\pi}{2}} \sin^{2} x \, dx = \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos(2x)}{2} \, dx$

This can be separated into two integrals:

$= \frac{1}{2} \int_{0}^{\frac{\pi}{2}} 1 \, dx - \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos(2x) \, dx$

Calculating these integrals, we find:

$\int \cos(2x) \, dx = \frac{1}{2} \sin(2x)$

Evaluating from ( 0 ) to ( \frac{\pi}{2} ):

$= \frac{1}{2} \left[ \sin(\pi) - \sin(0) \right] = 0$

Putting it all together:

$= \frac{1}{2} \cdot \frac{\pi}{2} - 0 = \frac{\pi}{4}$

Thus, the exact value is:

$\int_{0}^{\frac{\pi}{2}} \sin^{2} x \, dx = \frac{\pi}{4}$