Class 12 Integrals · Long Answer3 Marksdefinite_integration
Class 12 Maths · Integrals · Long Answer #172
Evaluate the convergent improper integral \(\int_0^{\pi/2}\ln(\sin x)dx\).
✅ Step-by-Step Solution
Final Answer: −(π/2)ln2
- Let I be the integral. Reflection gives I=∫₀^(π/2)ln(cos x)dx; both logarithmic endpoint singularities are integrable.
- Add: 2I=∫₀^(π/2)ln(sin2x)dx−(π/2)ln2.
- With u=2x, the first integral is (1/2)∫₀^πln(sin u)du=I by symmetry about π/2. Thus I=−(π/2)ln2.
- Therefore the integral equals −(π/2)ln2.
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] Let I be the integral. Reflection gives I=∫₀^(π/2)ln(cos x)dx; both logarithmic endpoint singularities are integrable.
- [1 mark] Add: 2I=∫₀^(π/2)ln(sin2x)dx−(π/2)ln2.
- [1 mark] With u=2x, the first integral is (1/2)∫₀^πln(sin u)du=I by symmetry about π/2. Thus I=−(π/2)ln2.
- [0.5 marks] Therefore the integral equals −(π/2)ln2.
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