Advertisements
Advertisements
प्रश्न
Integrate the function `(sec^2 x)/sqrt(tan^2 x + 4)`
Advertisements
उत्तर
Let `I = int (sec^2 x)/sqrt(tan^2 x + 4) dx`
Put tan x = t
sec2 x dx = dt
Hence, `I = int dt/sqrt(t^2 + 4) dt`
`= log abs ((t + sqrt(t^2 + 4))+ C` `....[∵ int dx/sqrt (a^2 + x^2) = log |x + sqrt(x^2 + a^2)| + C]`
`= log abs (tan x + sqrt(tan^2 x + 4)) + C`
APPEARS IN
संबंधित प्रश्न
Evaluate: `int(5x-2)/(1+2x+3x^2)dx`
Evaluate : ` int x^2/((x^2+4)(x^2+9))dx`
find : `int(3x+1)sqrt(4-3x-2x^2)dx`
Find:
`int(x^3-1)/(x^3+x)dx`
Integrate the function `(3x^2)/(x^6 + 1)`
Integrate the function `1/sqrt(1+4x^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/sqrt(x^2 +2x + 2)`
Integrate the function `1/sqrt(7 - 6x - x^2)`
Integrate the function `1/sqrt((x -1)(x - 2))`
Integrate the function `1/sqrt(8+3x - x^2)`
Integrate the function `(x + 2)/sqrt(x^2 -1)`
Integrate the function `(x + 2)/sqrt(4x - x^2)`
Integrate the function `(x + 3)/(x^2 - 2x - 5)`
`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(1+ 3x - x^2)`
Integrate the function:
`sqrt(1+ x^2/9)`
`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
If θ f(x) = `int_0^x t sin t dt` then `f^1(x)` is
`int (a^x - b^x)^2/(a^xb^x)dx` equals ______.
