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Evaluate the following : ∫0π4cos2x1+cos2x+sin2x⋅dx - Mathematics and Statistics

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Sum

Evaluate the following : `int_0^(pi/4) (cos2x)/(1 + cos 2x + sin 2x)*dx`

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Solution

Let I = `int_0^(pi/4) (cos2x)/(1 + cos 2x + sin 2x)*dx`

= `int_0^(pi/4) (cos^2x - sin^2x)/(2cos^2x + 2 sin x cosx)*dx`

= `int_0^(pi/4) ((cosx - sinx)(cosx + sinx))/(2cosx(cosx + sinx))*dx`

= `int_0^(pi/4) (cosx - sinx)/(2cosx)*dx`

= `(1)/(2) int_0^(pi/4) [cosx/cosx - sinx/cosx]*dx`

= `(1)/(2) [int_0^(pi/4) 1*dx - int_0^(pi/4) tanx*dx]`

= `(1)/(2){[x]_0^(pi/4) - [log (sec x)]_0^(pi/4)}`

= `(1)/(2)[(pi/4 - 0) - (log sec  pi/4 - log sec 0)]`

= `(1)/(2)[pi/4 - log sqrt(2) + log 1]`

= `(1)/(2)[pi/4 - log sqrt(2)]`.                    ...[∵ log 1 = 0]

Concept: Fundamental Theorem of Integral Calculus
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APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 4 Definite Integration
Miscellaneous Exercise 4 | Q 3.06 | Page 176
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