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Evaluate: ∫0π2cosx(1+sinx)(2+sinx) dx

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प्रश्न

Evaluate: `int_0^(pi/2) cos x/((1 + sinx)(2 + sinx))  "d"x`

योग
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उत्तर

Let I = `int_0^(pi/2) cos x/((1 + sinx)(2 + sinx))  "d"x`

Put sin x = t

∴ cos x dx = dt

When x = 0, t = 0 and when x = `pi/2`, t = 1

∴ I = `int_0^1 "dt"/((1 + "t")(2 + "t"))`

Let `1/((1 + "t")(2 + "t")) = "A"/(1 + "t") + "B"/(2 + "t")` ........(i)

∴ 1 = A(2 + t) + B(1 + t)  ........(ii)

Putting t = −1 in (ii), we get

A = 1

Putting t = −2 in (ii), we get

1 = − B

∴ B = −1

 From (i), we get

`1/((1 + "t")(2 + "t")) = 1/(1 + "t") - 1/(2 + "t")`

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

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

= `[log|1 + "t"|]_0^1 - [log|2 + "t"|]_0^1`

= (log 2 − log 1) − (log 3 − log 2)

= `log 2 - 0 - log(3/2)`

= `log(2 xx 2/3)`

∴ I = `log(4/3)`

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Methods of Evaluation and Properties of Definite Integral
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2.4: Definite Integration - Short Answers II

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