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

Sum

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

#### 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
Is there an error in this question or solution?
Chapter 4: Definite Integration - Miscellaneous Exercise 4 [Page 176]

#### 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
Share