हिंदी

Π ∫ 0 √ 1 − X 1 + X D X =π 2,π 2 − 1,π 2 + 1, π + 1

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प्रश्न

`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`

विकल्प

  • \[\frac{\pi}{2}\]
  • \[\frac{\pi}{2} - 1\]

  • \[\frac{\pi}{2} + 1\]
  •  π + 1

  • None of these

MCQ
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उत्तर

 None of the given option is correct.

\[\text{We have}, \]

\[I = \int_0^1 \sqrt{\frac{1 - x}{1 + x}} d x\]

`int_0^1 sqrt((1 - "x")/(1 + "x") xx (1 - "x")/(1 - "x"))   "dx"`

\[ = \int_0^1 \frac{1 - x}{\sqrt{1 - x^2}}dx\]

\[ = \int_0^1 \frac{1}{\sqrt{1 - x^2}}dx - \int_0^1 \frac{x}{\sqrt{1 - x^2}}dx\]

`=> int_0^1 1/sqrt(1 - "x"^2)  "dx" - int_0^1 "x"/sqrt(1 - "x"^2)`dx     ......`[int 1/(sqrt ("a"^2 - "x"^2)) "dx" = "sin"^-1 "x"/"a" + "C"]`

`= > ["sin"^-1 "x"/1]_0^1 + int_1^0 1/sqrt"t"  "dt"/2`

`=> [sin^-1 (1) - sin^-1(0)] + 1/2 int_1^0 "t"^(-1/2)`dt

 `=> pi/2 - 0 + 1/2 [2"t"^(1/2)]_0^1`

`=> pi/2 + (1 - "x"^2)^(1/2)int_0^1`

`=> pi/2 + [(1 - 1)^(1/2) - (1 - 0)^(1/2)]`

`=> pi/2 - 1^(1/2)`

`=> (pi/2 - 1)`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - MCQ [पृष्ठ ११७]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
MCQ | Q 10 | पृष्ठ ११७

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