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Evaluate: ∫0π2 sin2x1+ sin2x dx

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प्रश्न

Evaluate: `int_0^(pi/2)  (sin2x)/(1 +  sin^2x)  "d"x`

योग
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उत्तर

`int_0^(pi/2)  (sin2x)/(1 +  sin^2x)  "d"x = [log|1 + sin^2x|]_0^(pi/2)`      .......`[∵ int ("f'"(x))/("f"(x))  "d"x = log|"f"(x)| + "c"]`

= `log  |1 + sin^2(pi/2)| - log|1 + sin^2 0|`

= log |1 + 1| – log 1

= log 2 – 0

= log 2

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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 - Very Short Answers

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