JEEAdvanced
Evaluate I = \int_0^{2\pi} \frac{dx}{(5-4\cos x)^2}. The result is…
Question
Evaluate I = \int_0^{2\pi} \frac{dx}{(5-4\cos x)^2}. The result is \frac{m}{n}\pi with m/n in lowest terms. Find m+n.
✓ Verified answer: 37checked by our engine — not a guess
Step-by-step solution
Start from the standard result F(a)=\int_0^{2\pi}\frac{dx}{a-4\cos x}=\frac{2\pi}{\sqrt{a^2-16}} for a>4 (Weierstrass t=\tan(x/2) substitution).
Differentiate both sides with respect to a: \frac{dF}{da} = -\int_0^{2\pi}\frac{dx}{(a-4\cos x)^2} = \frac{d}{da}\Big(2\pi (a^2-16)^{-1/2}\Big) = -2\pi a (a^2-16)^{-3/2}.
Hence \int_0^{2\pi}\frac{dx}{(a-4\cos x)^2}=\frac{2\pi a}{(a^2-16)^{3/2}}.
Put a=5: (25-16)^{3/2}=9^{3/2}=27, giving I = \frac{2\pi\cdot5}{27} = \frac{10\pi}{27}.
So m=10, n=27 and m+n = 37.
Final answer37
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 · MathEvaluate I = \int_0^{\pi/2} \ln^2(\sin x)\,dx and give its value…JEE · MathUsing the symmetry x\to\pi-x, evaluate I = \int_0^{\pi} x\,\sin^4…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