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Evaluate the following : ∫1cos2x+3sin2x.dx - Mathematics and Statistics

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प्रश्न

Evaluate the following : `int (1)/(cos2x + 3sin^2x).dx`

योग

उत्तर

Let I = `int (1)/(cos2x + 3sin^2x).dx`

= `int (1)/(1 - 2sin^2x + 3sin^2x).dx`

= `int(1)/(1 + sin^2x).dx`

Dividing both numerator and denominator by cos2x, we get

I = `int(sec^2x dx)/(sec^2x + tan^2x)`

= `int (sec^2x dx)/(1 + tan^2x + tan^2x)`

= `int (sec^2x dx)/(2tan^2x + 1)`

Put tan x = t
∴ sec2x dx = dt

∴ I = `int (1)/(2t^2 + 1)dt`

= `(1)/(2) int (1)/(t^2 + (1/sqrt(2))^2)dt`

= `(1)/(2) xx (1)/((1/sqrt(2)))tan^-1 (t/(1/sqrt(2))) + c`

= `(1)/sqrt(2)tan^-1 (sqrt(2)tan x) + c`.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Indefinite Integration - Exercise 3.2 (B) [पृष्ठ १२३]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board
अध्याय 3 Indefinite Integration
Exercise 3.2 (B) | Q 1.19 | पृष्ठ १२३

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