Advertisements
Advertisements
प्रश्न
Find `int (dx)/sqrt(4x - x^2)`
Advertisements
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
find : `int(3x+1)sqrt(4-3x-2x^2)dx`
Integrate the function `1/sqrt((2-x)^2 + 1)`
Integrate the function `1/sqrt(9 - 25x^2)`
Integrate the function `(x - 1)/sqrt(x^2 - 1)`
Integrate the function `x^2/sqrt(x^6 + a^6)`
Integrate the function `1/sqrt(x^2 +2x + 2)`
Integrate the function `1/(9x^2 + 6x + 5)`
Integrate the function `1/sqrt((x -1)(x - 2))`
Integrate the function `(5x - 2)/(1 + 2x + 3x^2)`
Integrate the function `(6x + 7)/sqrt((x - 5)(x - 4))`
`int dx/(x^2 + 2x + 2)` equals:
`int dx/sqrt(9x - 4x^2)` equals:
Integrate the function:
`sqrt(x^2 + 4x +1)`
Integrate the function:
`sqrt(x^2 + 4x - 5)`
Integrate the function:
`sqrt(x^2 + 3x)`
Find `int dx/(5 - 8x - x^2)`
Integration of \[\frac{1}{1 + \left( \log_e x \right)^2}\] with respect to loge x is
\[\int\frac{8x + 13}{\sqrt{4x + 7}} \text{ dx }\]
