Advertisements
Advertisements
प्रश्न
Integrate the function:
`sqrt(1-4x - x^2)`
Advertisements
उत्तर
Let `I = sqrt(1 - 4x - x^2)`
`= int sqrt(1 - (x^2 + 4x + 4) + 4) dx`
`= int sqrt(5 - (x + 2)^2) dx`
`= int sqrt ((5)^2 - (x+2)^2) dx`
Now, `[because int sqrt (a^2 - x^2)dx = x/2 sqrt (a^2 - x^2) + a^2/2 sin^-1 x/a+C]`
Here, on putting 5 in place of a2 and (x + 2) in place of x,
I = `1/2 (x + 2) sqrt(5 + (x + 2)^2) + 5/2 sin^-1 (x + 2)/sqrt5 + C`
`= 1/2 (x + 2) sqrt(1 - 4x - x^2) + 5/2 sin^-1 (x + 2)/sqrt5 + 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 `(3x)/(1+ 2x^4)`
Integrate the function `x^2/(1 - x^6)`
Integrate the function `x^2/sqrt(x^6 + a^6)`
Integrate the function `1/sqrt((x - a)(x - b))`
Integrate the function `(x + 2)/sqrt(4x - x^2)`
`int dx/(x^2 + 2x + 2)` equals:
`int dx/sqrt(9x - 4x^2)` equals:
Integrate the function:
`sqrt(1- 4x^2)`
Integrate the function:
`sqrt(x^2 + 4x - 5)`
Integrate the function:
`sqrt(x^2 + 3x)`
Integrate the function:
`sqrt(1+ x^2/9)`
`int sqrt(1+ x^2) dx` is equal to ______.
`int sqrt(x^2 - 8x + 7) dx` is equal to ______.
Integration of \[\frac{1}{1 + \left( \log_e x \right)^2}\] with respect to loge x is
Find : \[\int\left( 2x + 5 \right)\sqrt{10 - 4x - 3 x^2}dx\] .
Find:
`int_(-pi/4)^0 (1+tan"x")/(1-tan"x") "dx"`
`int (a^x - b^x)^2/(a^xb^x)dx` equals ______.
