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JEE Mains · Maths · STD 12 - 7.2 definite integral

સંકલિત \(\int_0^\pi \frac{8 x d x}{4 \cos ^2 x+\sin ^2 x}\) = ___

  1. A \(2 \pi^2\)
  2. B \(4 \pi^2\)
  3. C \(\pi^2\)
  4. D \(\frac{3 \pi^2}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(2 \pi^2\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & I=\int_0^\pi \frac{8 x d x}{4 \cos ^2 x+\sin ^2 x} \\ & I=\int_0^\pi \frac{8(\pi-x) d x}{4 \cos ^2 x+\sin ^2 x} \end{aligned}\) \(2 \mathrm{I}=8 \pi \int_0^\pi \frac{d x}{4 \cos ^2 x+\sin ^2 x}\)…
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