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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Evaluate : ∫-π4π411-sinx⋅dx - Mathematics and Statistics

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प्रश्न

Evaluate:

`int_(-pi/4)^(pi/4) (1)/(1 - sinx)*dx`

मूल्यांकन

उत्तर

`int_(-pi/4)^(pi/4) (1)/(1 - sinx)*dx`

= `int_(-pi/4)^(pi/4) (1)/(1 - sinx)*(1 + sinx)/(1 + sinx)*dx`

= `int_(-pi/4)^(pi/4)(1 + sinx)/(1 - sin^2x)*dx`

= `int_(-pi/4)^(pi/4)(1 + sinx)/(cos^2x)*dx`

= `int_(-pi/4)^(pi/4) 1/(cos^2x) dx + int_(-pi/4)^(pi/4) (sinx)/(cos^2x)*dx`

= `int_(-pi/4)^(pi/4)sec^2x  dx + int_(-pi/4)^(pi/4) 1/cosx*sinx/cosxdx`

= `[tanx]_(-pi/4)^(pi/4) + int_(-pi/4)^(pi/4) secx tanx*dx`

= `[tan  pi/4 - tan  (-pi/4)] + [secx]_(-pi/4)^(pi/4)`

= `[1 + 1] + [sec  pi/4 - sec  ((-pi)/4)]`

= `2 + [sqrt2 - sqrt2]`

= 2 + 0

I = 2

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Fundamental Theorem of Integral Calculus
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Definite Integration - Exercise 4.2 [पृष्ठ १७१]

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