English

Evaluate the following : ∫π4π2cosθ[cos θ2+sin θ2]3⋅dθ - Mathematics and Statistics

Advertisements
Advertisements

Question

Evaluate the following : `int_(pi/4)^(pi/2) (cos theta)/[cos  theta/2 + sin  theta/2]^3*d theta`

Sum
Advertisements

Solution

Let I = `int_(pi/4)^(pi/2) (cos theta)/[cos  theta/2 + sin  theta/2]^3*d theta`

= `int_(pi/4)^(pi/2) (cos^2  theta/2 - sin^2  theta/2)/[cos  theta/2 + sin  theta/2]^3*d theta`

= `int_(pi/4)^(pi/2)((cos  theta/2 - sin  theta/2)(cos  theta/2 + sin  theta/2))/[cos  theta/2 + sin  theta/2]^3*d theta`

= `int_(pi/4)^(pi/2) (cos  theta/2 - sin  theta/2)/[cos  theta/2 + sin  theta/2]^2*d theta`

Put `cos  theta/2 - sin  theta/2` = t

∴ `(-1/2 sin  theta/2 +1/2 cos  theta/2)*d theta` = dt

∴ `(cos  theta/2 - sin  theta/2)*d theta = 2*dt`

When θ = `pi/(4), t = cos  pi/(8) + sin  pi/(8)`

When θ = `pi/(2), t = cos  pi/(4) + sin  pi/(4) = (1)/sqrt(2) + (1)/sqrt(2) = sqrt(2)`

∴  I = `int_(cos  pi/8 + sin  pi/8)^sqrt(2) (1)/t^2* 2dt`

= `2 int_(cos  pi/8 + sin  pi/8)^sqrt(2) t^-2*dt`

= `2[(t^-1)/-1]_(cos  pi/8 + sin  pi/8)^sqrt(2)`

= `[(-2)/t]_(cos  pi/8 + sin  pi/8)^sqrt(2)`

= `- (2)/sqrt(2) + (2)/(cos  pi/8 + sin  pi/8)`

= `(2)/(cos  pi/8 + sin  pi/8) - sqrt(2)`.

shaalaa.com
  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) [English] Standard 12 Maharashtra State Board
Chapter 4 Definite Integration
Miscellaneous Exercise 4 | Q 2.02 | Page 176
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×