JEE Main 2025April, Unspecified (session-level dataset)MathematicsMediumIntegral Calculus369× since 2002The integral ∫0π8x4cos2x+sin2x dx\displaystyle\int_{0}^{\pi}\frac{8x}{4\cos^2x + \sin^2x}\,dx∫0π4cos2x+sin2x8xdx is equal to:AOption A. 2π22\pi^22π2BOption B. 4π24\pi^24π2COption C. π2\pi^2π2DOption D. 3π22\tfrac{3\pi^2}{2}23π2CheckSkip · show solution
