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$$\rm \displaystyle \int_{0}^{\dfrac{\pi}{2}} \dfrac{cos^2x dx}{cos^2x+4sin^2x}$$

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$$\rm \displaystyle \int_{0}^{\dfrac{\pi}{2}} \dfrac{cos^2x dx}{cos^2x+4sin^2x}$$

Posted by ⠀ 3 years, 6 months ago

CBSE > Class 12 > Mathematics

1 answers

Hitkar Miglani 3 years, 5 months ago

Ans = π / 6

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$$\rm \displaystyle \int_{0}^{\dfrac{\pi}{2}} \dfrac{cos^2x dx}{cos^2x+4sin^2x}$$

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