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प्रश्न
`int_0^(pi/4) (cos^2 x)/(cos^2 x + 4 sin^2 x) dx` =
पर्याय
`pi/4 + 2/3 tan^-1 3`
`- pi/3 - 2/3 tan^-1 3`
`- pi/12 + 2/3 tan^-1 2`
`pi/6 - 2/3 tan^-1 4`
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उत्तर
`bb(- pi/12 + 2/3 tan^-1 2)`
Explanation:
We have, `int_0^(pi/4) (cos^2 x)/(cos^2 x + 4 sin^2 x) dx`
= `int_0^(pi/4) (dx)/(1 + 4 tan^2 x)`
Let tan x = t
= sec2 x dx = dt
= `dx = (dx)/(sec^2 x) = (dt)/(1 + t^2)` ...(i)
= when x = 0, t = 0, and when x = `pi/4`, t = 1
From Eq. (i), `int_0^1 (dt)/((1 + t^2)(1 + 4t^2))`
= `1/3 int_0^1 (4(1 + t^2) - (1 + 4t^2))/((1 + 4t^2)(1 + t^2))dx`
= `1/3 int_0^1 (4/(1 + 4t^2) - 1/(1 + t^2)dt)`
= `1/3 [4/4 int_0^1 (dt)/((1/2)^2 + t^2) - int_0^1 1/(1 + t^2)dt]`
= `1/3 [1/(1"/"2) (tan^-1 t/(1"/"2))_0^1 - (tan^-1 t)_0^1]`
= `1/3 [2 (tan^-1 2t)_0^1 - pi/4]`
= `1/3 [2 tan^-1 (2) - pi/4]`
= `-pi/12 + 2/3 tan^-1 2`
