Advertisements
Advertisements
प्रश्न
`int dx/sqrt(9x - 4x^2)` equals:
विकल्प
`1/9 sin^-1 ((9x - 8)/8) + C`
`1/2 sin^-1 ((8x - 9)/9) + C`
`1/3 sin^-1 ((9x - 8)/8) + C`
`1/2 sin^-1 ((9x - 8)/9) + C`
Advertisements
उत्तर
`1/2 sin^-1 ((8x - 9)/9) + C`
Explanaton:
Let `I = int dx/sqrt(9x^2 - 4x^2)`
`= 1/2 int dx/sqrt(9/4 x - x^2)`
`= 1/2 int dx/(sqrt(- (x^2 - 9/4 x)))`
`= 1/2 dx/sqrt(- (x^2 - 9/4 + 81/64) + 81/64)`
`= 1/2 int dx/sqrt(81/64 - (x - 9/8)^2) ... [because int 1/(a^2 - x^2) dx = sin^-1 x/a]`
`= 1/2 sin^-1 ((x - 9/8)/(9/8)) +C`
`= 1/2 sin^-1 ((8x - 9)/9) + C`
APPEARS IN
संबंधित प्रश्न
Evaluate: `int(5x-2)/(1+2x+3x^2)dx`
Find:
`int(x^3-1)/(x^3+x)dx`
Integrate the function `(3x^2)/(x^6 + 1)`
Integrate the function `1/sqrt(9 - 25x^2)`
Integrate the function `(3x)/(1+ 2x^4)`
Integrate the function `x^2/(1 - x^6)`
Integrate the function `(x - 1)/sqrt(x^2 - 1)`
Integrate the function `1/(9x^2 + 6x + 5)`
Integrate the function `1/sqrt((x -1)(x - 2))`
Integrate the function `1/sqrt(8+3x - x^2)`
Integrate the function `1/sqrt((x - a)(x - b))`
Integrate the function `(x + 2)/sqrt(x^2 -1)`
Integrate the function `(x + 2)/sqrt(4x - x^2)`
Integrate the function `(5x + 3)/sqrt(x^2 + 4x + 10)`
`int dx/(x^2 + 2x + 2)` equals:
Integrate the function:
`sqrt(4 - x^2)`
Integrate the function:
`sqrt(x^2 + 4x + 6)`
Integrate the function:
`sqrt(x^2 + 4x +1)`
Integrate the function:
`sqrt(1-4x - x^2)`
Integrate the function:
`sqrt(x^2 + 4x - 5)`
Integrate the function:
`sqrt(1+ 3x - x^2)`
Integrate the function:
`sqrt(x^2 + 3x)`
Integrate the function:
`sqrt(1+ x^2/9)`
\[\int\frac{8x + 13}{\sqrt{4x + 7}} \text{ dx }\]
\[\int\frac{1 + x + x^2}{x^2 \left( 1 + x \right)} \text{ dx}\]
Find:
`int_(-pi/4)^0 (1+tan"x")/(1-tan"x") "dx"`
Find `int (dx)/sqrt(4x - x^2)`
Find: `int (dx)/(x^2 - 6x + 13)`
`int (a^x - b^x)^2/(a^xb^x)dx` equals ______.
