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प्रश्न
Evaluate the definite integral:
`int_0^(pi/2) (cos^2 x dx)/(cos^2 x + 4 sin^2 x)`
उत्तर
Let `I = int_0^(pi/2) (cos^2 x )/(cos^2 x + 4 sin^2 x)`dx
`int_0^(pi/2) (cos^2 x)/(cos^2 x + 4(1 - cos^2 x))`dx
`= int_0^(pi/2) (cos^2x)/(4 - 3 cos^2 x)`dx
`= - 1/3 int_0^(pi//2) (4 - 3 cos^2 x - 4)/(4 - 3 cos^2 x)`dx
`= - 1/3 int_0^(pi/2) (1 - 4/(4 - 3 cos^2 x))`dx
`= - 1/3 int_0^(pi/2) 1 * dx + 4/3 int_0^(pi/2) dx/(4 - 3 cos^2 x)`
`= - 1/3 (pi/2) + 4/3 int_0^(pi/2) (sec^2x)/(4 sec^2 x - 3)`dx
`= - pi/6 + 4/3 int_0^(pi/2) (sec^2 x)/(4 (1 + tan^2 x - 3))`dx
⇒ Put tan x = t
sec2 x dx = dt
When x = 0, t = 0 and when x = `pi/2, t = oo`
I = `- pi/6 + 4/3 int_0^oo dt/(4(1 + t^2) - 3)`
`= pi/6 + 4/3 int_0^oo dt/(4t^2 + 1)`
`= - pi/6 + 4/3 * 1/4 int_0^oo dt/(t^2 + 1/4)`
`= - pi/6 + 1/3 * 2 [tan^-1 t/(1//2)]_0^oo`
`= - pi/6 + 2/3 * [tan^-1 2t]_0^oo`
`= - pi/6 + 2/3 [tan^-1 oo - tan^-1 0]`
`= - pi/6 + 2/3 * [pi/2 - 0]`
`= - pi/6 + pi/3`
`= pi/6`
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