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

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प्रश्न

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

योग

उत्तर

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)`.

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Fundamental Theorem of Integral Calculus
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Definite Integration - Miscellaneous Exercise 4 [पृष्ठ १७६]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board
अध्याय 4 Definite Integration
Miscellaneous Exercise 4 | Q 3.04 | पृष्ठ १७६

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