The integral ∫₀^π (8 x d x)/(4 cos ² x+sin ² x) is equal to

Solution

Maths sir · 4 steps

An x on top with a denominator symmetric about π/2? Replace x by π − x and the x disappears.

1. Use the King property. Replace x by π - x. Since cos²(π - x) = cos² x and sin²(π - x) = sin² x, the denominator does not change, so I = ∫₀^π 8(π - x)/(4 cos² x + sin² x) dx.
2. Add the two forms. Adding this to the original gives 2I = 8π ∫₀^π dx/(4 cos² x + sin² x). The integrand repeats symmetrically about π/2, so ∫₀^π = 2∫₀^(π/2), and I = 8π ∫₀^(π/2) dx/(4 cos² x + sin² x).
3. Divide by cos squared. Divide top and bottom by cos² x and put t = tan x, so dt = sec² x dx. As x goes from 0 to π/2, t goes from 0 to ∞.

   ∫₀^(π/2) (sec² x dx)/(4 + tan² x) = ∫₀^∞ dt/(4 + t²) = [1/2 tan⁻¹ t/2]₀^∞ = π/4

4. So. I = 8π × π/4 = 2π², which is option (a).

The idea to keep. When x multiplies a function that is unchanged under x → a + b − x, add the two forms of I. The x drops out and you only integrate the plain function.

Answer from the NTA final answer key. Solution by PrepWiser.