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Evaluate : ∫121ecos-1xsin-1x1-x2⋅dx - Mathematics and Statistics

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प्रश्न

Evaluate : `int _((1)/(sqrt(2)))^1 (e^(cos^-1x) sin^-1x)/(sqrt(1 - x^2))*dx`

योग

उत्तर

Let I = `int _((1)/(sqrt(2)))^1 (e^(cos^-1x) sin^-1x)/(sqrt(1 - x^2))*dx`

Put sin–1 x = t

∴ `(1)/sqrt(1 - x^2)*dx` = dt

When x = 1, t = `sin^-1 1 = pi/(2)`

When x = `1/sqrt(2), t = sin^-1 1/sqrt(2) = pi/(4)`

Also, `cos^-1 x = pi/2 - sin^-1x = pi/(2) - t`

∴ I = `int_(i/4)^(pi/2) e^(pi/2 - t)*t  dt`

= `e^(pi/2) int_(i/4)^(pi/2) te^-t dt`

= `e^(pi/2) {[t int e^-t dt]_(pi/4)^(pi/2) - int_(i/4)^(pi/2)[d/dt (t) int e^-t dt]*dt}`

= `e^(pi/2){[ - te^-t]_(pi/4)^(pi/2)  - int_(i/4)^(pi/2) (1)( - e^-t)*dt}`

= `e^(pi/2) {(-pi)/(2) e^(-pi/2) + pi/(4) e^(-pi/4) + int_(i/4)^(pi/2) e^-t *dt}`

= `- pi/(2) e^o + pi/(4) e^(pi/2 - pi/4) + e^(pi/2)[- e^-t]^(pi/2)`

= `- pi/(2) + pi/(4) e^(pi/4) + e^(pi/2)[ - e^(-pi/2) + e^((-pi)/4)]`

= `- pi/(2) + e^(pi/4) pi/(4) - e^o + ^(pi/2 - pi/4)`

= `- pi/(2) + e^(pi/4) pi/(4) - 1 + e^(pi/4)`

= `e^(pi/4) (pi/4 + 1) - (pi/2 + 1)`.

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Fundamental Theorem of Integral Calculus
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Definite Integration - Exercise 4.2 [पृष्ठ १७२]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board
अध्याय 4 Definite Integration
Exercise 4.2 | Q 2.14 | पृष्ठ १७२

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