Class 12 Integrals · Very Long Answer5 Marksdefinite_integration
Class 12 Maths · Integrals · Very Long Answer #175
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. Near 0, ln(sin x)=ln x+O(x²), so the endpoint singularity is integrable.
- Reflection x↦π/2−x gives I=∫₀^(π/2)ln(cos x)dx. Add the two equal integrals.
- Using sin x cos x=sin2x/2 yields 2I=J−(π/2)ln2, where J=∫₀^(π/2)ln(sin2x)dx.
- Set u=2x. Then J=(1/2)∫₀^πln(sin u)du=I, by symmetry about π/2. Solve 2I=I−(π/2)ln2.
- Therefore the integral equals −(π/2)ln2.
📋 CBSE Stepwise Marking Scheme
- [1 mark] Let I be the integral. Near 0, ln(sin x)=ln x+O(x²), so the endpoint singularity is integrable.
- [1 mark] Reflection x↦π/2−x gives I=∫₀^(π/2)ln(cos x)dx. Add the two equal integrals.
- [1 mark] Using sin x cos x=sin2x/2 yields 2I=J−(π/2)ln2, where J=∫₀^(π/2)ln(sin2x)dx.
- [1 mark] Set u=2x. Then J=(1/2)∫₀^πln(sin u)du=I, by symmetry about π/2. Solve 2I=I−(π/2)ln2.
- [1 mark] Therefore the integral equals −(π/2)ln2.
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