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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Evaluate: ∫0π4 cos2x1+cos2x+sin2x dx - Mathematics and Statistics

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प्रश्न

Evaluate: `int_0^(pi/4)  (cos2x)/(1 + cos 2x + sin 2x)  "d"x`

बेरीज
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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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Methods of Evaluation and Properties of Definite Integral
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पाठ 2.4: Definite Integration - Long Answers III

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