# Evaluate the following : ∫π53π10sinxsinx+cosx⋅dx - Mathematics and Statistics

Sum

Evaluate the following : int_(pi/5)^((3pi)/10) sinx/(sinx + cosx)*dx

#### Solution

Let I = int_(pi/5)^((3pi)/10) sinx/(sinx + cosx)*dx    ...(1)

We use the property, int_a^b f(x)*dx = int_a^bf(a + b - x)*dx.

Here a = pi/(5), b = (3pi)/(10).

Hence changing x by pi/(5) + (3pi)/(10) - x, we get,

I = int_(pi/5)^((3pi)/(10)) (sin(pi/5 + (3pi)/(10) - x))/(sin(pi/5 + (pi)/(10) - x) + cos(pi/5 + (3pi)/(10) - x))*dx

= int_(pi/5)^((3pi)/(10)) (sin(pi/2 - x))/(sin(pi/2 - x) + cos(pi/2 - x))*dx

= int_(pi/5)^((3pi)/(10)) cosx/(cosx + sinx)*dx      ...(2)

Adding (1) and (2), we get,

2I = int_(pi/5)^((3pi)/(10)) sinx/(sinx + cosx)*dx + int_(pi/5)^((3pi)/(10)) cosx/(cosx + sinx)*dx

= int_(pi/5)^((3pi)/(10)) (sinx + cosx)/(sinx + cosx)*dx

= int_(pi/5)^((3pi)/(10))1*dx = [x]_(pi/5)^((3pi)/10)

= (3pi)/(10) - pi/(5)

= pi/(10)

∴ I = pi/(20).

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 3.04 | Page 176
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