JEEOlympiad
Evaluate I = \int_0^{\pi/2} \ln^2(\sin x)\,dx and give its value…
Question
Evaluate I = \int_0^{\pi/2} \ln^2(\sin x)\,dx and give its value rounded to 4 decimal places.
✓ Verified answer: 2.0466checked by our engine — not a guess
Step-by-step solution
Use the Fourier/Beta-function approach.
Consider J(s)=\int_0^{\pi/2}\sin^s x\,dx = \frac{\sqrt\pi}{2}\frac{\Gamma((s+1)/2)}{\Gamma((s+2)/2)}.
Then I = \frac{d^2}{ds^2}J(s)\big|_{s=0}, since differentiating \sin^s x twice in s brings down \ln^2(\sin x).
Differentiating the Gamma expression and evaluating at s=0 (using \psi(1/2)=-\gamma-2\ln2, \psi(1)=-\gamma and the trigamma values \psi'(1/2)=\pi^2/2, \psi'(1)=\pi^2/6) yields the classical closed form I = \frac{\pi}{2}\Big(\ln^2 2 + \frac{\pi^2}{12}\Big).
Numerically \ln^2 2 = 0.480453, \pi^2/12 = 0.822467, sum 1.302920, times \pi/2 = 2.04662.
Final answer2.0466
Stuck on a problem like this?
Paste any JEE or NEET question — verified working, a confidence %, and an honest “not sure” instead of a bluff.
Solve my doubt →More Definite Integrals solutions
Evaluate I = \int_0^1 \frac{x^4(1-x)^4}{1+x^2}\,dx. It can be written…JEE · MathUsing the symmetry x\to\pi-x, evaluate I = \int_0^{\pi} x\,\sin^4…JEE · MathEvaluate I = \int_0^{2\pi} \frac{dx}{(5-4\cos x)^2}. The result is…JEE · MathBy differentiating under the integral sign, evaluate I =…JEE · MathUsing the reduction formula for powers of tangent, evaluate I =…JEE · MathUsing the Beta function (or Wallis reduction), evaluate I =…JEE · MathEvaluate I = \int_0^{1} \frac{(\ln x)^2}{1+x^2}\,dx. It equals…JEE · MathEvaluate I = \int_0^{\pi} \ln\!\big(5 - 4\cos x\big)\,dx and give its…JEE · Math