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Question 6
4. (i) Find \( \int \ln(x) \, dx \).\n\n(ii) Find the exact value of \( \int_{0}^{\frac{\pi}{2}} \sin^2(x) \, dx \).
Step 1
Answer
To find ( \int \ln(x) , dx ), we can use integration by parts. Let:
Using the integration by parts formula:
[ \int u , dv = uv - \int v , du ]
we get:
[ \int \ln(x) , dx = x \ln(x) - \int x \cdot \frac{1}{x} , dx = x \ln(x) - \int 1 , dx = x \ln(x) - x + C ]
Thus, ( \int \ln(x) , dx = x \ln(x) - x + C ).
Step 2
Answer
To find ( \int_{0}^{\frac{\pi}{2}} \sin^2(x) , dx ), we can use the identity:
[ \sin^2(x) = \frac{1}{2} (1 - \cos(2x)) ]
Therefore, the integral becomes:
[ \int_{0}^{\frac{\pi}{2}} \sin^2(x) , dx = \int_{0}^{\frac{\pi}{2}} \frac{1}{2} (1 - \cos(2x)) , dx ]
Splitting this into two parts:
[ = \frac{1}{2} \left( \int_{0}^{\frac{\pi}{2}} 1 , dx - \int_{0}^{\frac{\pi}{2}} \cos(2x) , dx \right) ]
Calculating each integral:
Combining these results:
[ = \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4} ]
Thus, ( \int_{0}^{\frac{\pi}{2}} \sin^2(x) , dx = \frac{\pi}{4} ).
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