Advertisements
Advertisements
प्रश्न
Integrate the function:
`sqrt(4 - x^2)`
Advertisements
उत्तर
Let `I = int sqrt (4 - x^2) dx`
`= int sqrt ((2)^2 - x^2) dx`
`= [x /2 sqrt ((2)^2 - x^2) + 4/2 sin^-1 (x/2)] + C` `...[int sqrt (a^2 - x^2) dx = x/2 sqrt (a^2 - x^2) + a^2/2 sin^-1 (x/a) + C]`
`= (x sqrt 4 - x^2)/2 + 4/2 sin^-1 (x/2) +C`
`(x sqrt(4 - x^2))/2 + 2 sin^-1 (x/2) + C`
APPEARS IN
संबंधित प्रश्न
Evaluate: `int(5x-2)/(1+2x+3x^2)dx`
find : `int(3x+1)sqrt(4-3x-2x^2)dx`
Find:
`int(x^3-1)/(x^3+x)dx`
Evaluate:
`int((x+3)e^x)/((x+5)^3)dx`
Integrate the function `1/sqrt(1+4x^2)`
Integrate the function `1/sqrt(x^2 +2x + 2)`
Integrate the function `1/sqrt(7 - 6x - x^2)`
Integrate the function `1/sqrt((x - a)(x - b))`
Integrate the function `(5x - 2)/(1 + 2x + 3x^2)`
Integrate the function `(x + 2)/sqrt(4x - x^2)`
Integrate the function `(x+2)/sqrt(x^2 + 2x + 3)`
Integrate the function `(5x + 3)/sqrt(x^2 + 4x + 10)`
Integrate the function:
`sqrt(1- 4x^2)`
Integrate the function:
`sqrt(x^2 + 4x + 6)`
Integrate the function:
`sqrt(1-4x - x^2)`
Integrate the function:
`sqrt(1+ 3x - x^2)`
`int sqrt(1+ x^2) dx` is equal to ______.
Evaluate : `int_2^3 3^x dx`
If θ f(x) = `int_0^x t sin t dt` then `f^1(x)` is
Find `int (dx)/sqrt(4x - x^2)`
What is a standard integral?
Evaluate \[\int \frac{dx}{x^2-a^2}\].
Evaluate \[\int \frac{dx}{\sqrt{a^2-x^2}}\].
Which trigonometric substitution is suitable for \[x^2+a^2\] and \[\sqrt{x^2+a^2}\]?
What is \[\frac{d}{dx}(ax^2+bx+c)\]?
Evaluate \[\int\frac{dx}{x^2-6x+13}\].
Evaluate \[\int\frac{dx}{3x^2+13x-10}\].
Evaluate \[\int\frac{x+2}{2x^2+6x+5}\,dx\].
Which statement is correct when integrating an expression that is not immediately a standard integral?
