Advertisements
Advertisements
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
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)`
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
The value of `int_2^3 x/(x^2 + 1)`dx is ______.
