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Class 12 Integrals · Very Long Answer5 Marksdefinite_integration

Class 12 Maths · Integrals · Very Long Answer #116

For a>0,b>0, evaluate \(\int_0^\pi x/[a^2\cos^2x+b^2\sin^2x]dx\).

✅ Step-by-Step Solution

Final Answer: π²/(2ab)
  1. Let g=1/(a²cos²x+b²sin²x), positive and symmetric under x↦π−x. Reflection gives I=(π/2)∫₀^πg(x)dx.
  2. The same symmetry gives ∫₀^πg=2∫₀^(π/2)g. Thus I=π∫₀^(π/2)dx/(a²cos²x+b²sin²x).
  3. Set u=b tan x. Then du=b sec²x dx, and dividing by cos²x makes the remaining integral (1/b)∫₀^∞du/(a²+u²).
  4. For a>0 this equals (1/(ab))[arctan(u/a)]₀^∞=π/(2ab); multiply by π.
  5. Therefore the integral equals π²/(2ab).

📋 CBSE Stepwise Marking Scheme

  • [1 mark] Let g=1/(a²cos²x+b²sin²x), positive and symmetric under x↦π−x. Reflection gives I=(π/2)∫₀^πg(x)dx.
  • [1 mark] The same symmetry gives ∫₀^πg=2∫₀^(π/2)g. Thus I=π∫₀^(π/2)dx/(a²cos²x+b²sin²x).
  • [1 mark] Set u=b tan x. Then du=b sec²x dx, and dividing by cos²x makes the remaining integral (1/b)∫₀^∞du/(a²+u²).
  • [1 mark] For a>0 this equals (1/(ab))[arctan(u/a)]₀^∞=π/(2ab); multiply by π.
  • [1 mark] Therefore the integral equals π²/(2ab).

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