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प्रश्न
Evaluate: `int_0^(pi/4) (cos2x)/(1 + cos 2x + sin 2x) "d"x`
उत्तर
Let I = `int_0^(pi/4) (cos2x)/(1 + cos 2x + sin 2x) "d"x`
= `int_0^(pi/4) (cos^2x - sin^2x)/(2cos^2x + 2sinx cosx) "d"x`
= `int_0^(pi/4) ((cosx + sinx)(cosx - sin x))/(2cos(cosx + sinx)) "d"x`
= `1/2 int_0^(pi/4) (cosx - sinx)/cosx "d"x`
= `1/2 int_0^(pi/4) (1 - tan x) "d"x`
= `1/2 int_0^(pi/4) "d"x - 1/2 int_0^(pi/4) tanx "d"x`
= `1/2[x]_0^(pi/4) - 1/2[log|sec x|]_0^(pi/4)`
= `1/2(pi/4 - 0) - 1/2[log|sec pi/4| - log|sec 0|]`
= `pi/8 - 1/2 (log sqrt(2) - log 1)`
= `pi/8 - 1/2 (log sqrt(2) - 0)`
∴ I = `1/2(pi/4 - log sqrt(2))`
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