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

Sum

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

#### 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
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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 3.05 | Page 176
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