# Evaluate the following : ∫0πx1+sin2x⋅dx - Mathematics and Statistics

Sum

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

#### Solution

Let I = int_0^pi x/(1 + sin^2x)*dx                     ...(1)

We use the property, int_^a f(x)*dx = int_0^a f(a - x)*dx

Here a = π.
Hence in I, changing x to π – x, we get

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

= int_0^pi (pi - x)/(1 + sin^2x)*dx

= int_0^pi pi/(1 + sin^2x)*dx

= -int_0^(pi) x/(1 + sin^2x)*dx

= int_0^(pi) pi/(1 sin^2x)*dx - "I"         ...[By (1)]

∴ 2I = pi int_0^(pi) 1/(1 + sin^2x)*dx
Dividing numerator and denominator by cos2x, we get

2I = pi int_0^(pi) (sec^2x)/(sec^2x + tan^2x)*dx

= pi int_0^(pi) (sec^2x)/(1 + 2tan^2x)*dx

Put tan x = t
∴ sec2x·dx = dt
When x = π, t = tan π = 0
When x = 0, t = tan 0 = 0

∴ 2I = pi int_0^(pi) dt/(1 + 2^2) = 0

∴ I = 0.                                    ...[∵ int_a^a f(x)*dx = 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 2.09 | Page 176
Share