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Evaluate: ∫0π4 cosx4-sin2x dx - Mathematics and Statistics

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प्रश्न

Evaluate: `int_0^(pi/4)  cosx/(4 - sin^2 x)  "d"x`

योग
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उत्तर

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

Put sin x = t

∴ cos x dx = dt

When x = 0, t = 0 and when x = `pi/4`,t  `1/sqrt(2)`

∴ I = `int_0^(1/sqrt(2)) "dt"/(4 - "t"^2)`

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

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

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

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

∴ I = `1/4 log((2sqrt(2) + 1)/(2sqrt(2) - 1))`

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Methods of Evaluation and Properties of Definite Integral
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अध्याय 2.4: Definite Integration - Short Answers I

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