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Evaluate : ∫0π4cosx4-sin2x⋅dx - Mathematics and Statistics

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प्रश्न

Evaluate : `int_0^(pi/4) (cosx)/(4 - sin^2x)*dx`

योग

उत्तर

Let I = `int_0^(pi/4) (cosx)/(4 - sin^2x)*dx`

Put sin x = t
∴ cos x·dx = dt

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

When x = 0, t = sin 0 = 0.

∴ I = `int_0^(1/sqrt(2)) dt/(2^2 - t^2)`

= `[1/(2(2)) log|(2 + t)/(2 - t)|]_0^((1)/sqrt(2))`

= `(1)/(4)[log((2 + 1/sqrt(2))/(2 - 1/sqrt(2))) - log((2 + 0)/(2  - 0))]`

= `(1)/(4)[log((2sqrt(2) + 1)/(2sqrt(2) - 1)) - log 1]`

= `(1)/(4)log((2sqrt(2) + 1)/(2sqrt(2) - 1))`.       ...[∵ log 1 = 0]

shaalaa.com
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 2.06 | पृष्ठ १७२

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