Advertisements
Advertisements
प्रश्न
Find `int (dx)/sqrt(4x - x^2)`
उत्तर
Let I = `int (dx)/sqrt(4x - x^2)`
= `int (dx)/sqrt(-(x^2 - 4x))`
= `int (dx)/sqrt(-(x^2 - 4x + 2^2 - 2^2))`
= `int (dx)/sqrt(-(x - 2)^2 - 2^2)`
= `int (dx)/sqrt(2^2 - (x - 2)^2)`
= `sin^-1 ((x - 2)/2) + C` ...`[∵ int (dx)/sqrt(a^2 - x^2) = sin^-1 (x/a) + C]`
APPEARS IN
संबंधित प्रश्न
Evaluate: `int(5x-2)/(1+2x+3x^2)dx`
Integrate the function `(x - 1)/sqrt(x^2 - 1)`
Integrate the function `x^2/sqrt(x^6 + a^6)`
Integrate the function `(sec^2 x)/sqrt(tan^2 x + 4)`
Integrate the function `1/sqrt(8+3x - x^2)`
Integrate the function `(4x+ 1)/sqrt(2x^2 + x - 3)`
Integrate the function `(5x - 2)/(1 + 2x + 3x^2)`
Integrate the function `(6x + 7)/sqrt((x - 5)(x - 4))`
Integrate the function `(x+2)/sqrt(x^2 + 2x + 3)`
`int dx/(x^2 + 2x + 2)` equals:
Integrate the function:
`sqrt(x^2 + 4x +1)`
Integrate the function:
`sqrt(1-4x - x^2)`
Integrate the function:
`sqrt(x^2 + 3x)`
Integrate the function:
`sqrt(1+ x^2/9)`
Find `int dx/(5 - 8x - x^2)`
Find `int (2x)/(x^2 + 1)(x^2 + 2)^2 dx`
Integration of \[\frac{1}{1 + \left( \log_e x \right)^2}\] with respect to loge x is
\[\int\frac{1 + x + x^2}{x^2 \left( 1 + x \right)} \text{ dx}\]
Find : \[\int\left( 2x + 5 \right)\sqrt{10 - 4x - 3 x^2}dx\] .
If θ f(x) = `int_0^x t sin t dt` then `f^1(x)` is
