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Question
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
Options
- \[\frac{\pi}{2}\]
\[\frac{\pi}{2} - 1\]
- \[\frac{\pi}{2} + 1\]
π + 1
None of these
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Solution
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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