Advertisements
Advertisements
प्रश्न
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
Advertisements
उत्तर
Let I = `int sqrt((1 + x)/(1 - x)) "d"x`
= `int 1/sqrt(1 - x^2) "d"x + int (x"d"x)/sqrt(1 - x^2)`
= `sin^-1x + 1`
When I1 = `(x"d"x)/sqrt(1 - x^2)`.
Put 1 – x2 = t2
⇒ –2x dx = 2t dt.
Therefore I1 = – dt = – t + C
= `- sqrt(1 - x^2) + "C"`
Hence I = `sin^-1x - sqrt(1 - x^2) + "C"`.
APPEARS IN
संबंधित प्रश्न
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
