Maharashtra State BoardHSC Arts 12th Board Exam
Advertisement Remove all ads

Evaluate the following : ∫01sin-1(2x1+x2)⋅dx - Mathematics and Statistics

Advertisement Remove all ads
Advertisement Remove all ads
Sum

Evaluate the following : `int_0^1 sin^-1 ((2x)/(1 + x^2))*dx`

Advertisement Remove all ads

Solution

Let I = `int_0^1 sin^-1 ((2x)/(1 + x^2))*dx`

Put x = tan t, i.e. t = tan–1x
∴ dx = sec2t·dt
When x = 0, t = tan–1 0 = 0

When x = 1, t = tan–1 = `pi/(4)`

∴ I = `int_0^(pi/4) sin^-1 ((2tant)/(1 + tan^2t))sec^2t*dt`

= `int_0^(pi/4) sin^-1 (sin2t) sec^2t*dt`

= `int_0^(pi/4) 2t sec^2t*dt`

= `[2t int sec^2t*dt]_0^(pi/4) - int_0^(pi/4) [d/dx (2t) int sec^2t*dt]`

= `[2t tan t]_0^(pi/4) - int_0^(pi/4) 2 tan t *dt`

= `[2* pi/4 tan  pi/4 - 0] - 2log(sec t)]_0^(pi/4)`

= `pi/2 - 2[log(sec  pi/4) - log (sec0)]`

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

= `pi/2 - 2[1/2 log 2 - 0]`

= `pi/(2) - log 2`.

Concept: Fundamental Theorem of Integral Calculus
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 4 Definite Integration
Miscellaneous Exercise 4 | Q 3.05 | Page 176
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×