Advertisements
Advertisements
प्रश्न
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
पर्याय
- \[\frac{\pi}{2}\]
\[\frac{\pi}{2} - 1\]
- \[\frac{\pi}{2} + 1\]
π + 1
None of these
Advertisements
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)`
APPEARS IN
संबंधित प्रश्न
If f(x) is a continuous function defined on [−a, a], then prove that
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
Choose the correct alternative:
If n > 0, then Γ(n) is
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Find `int sqrt(10 - 4x + 4x^2) "d"x`
