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
Integrate the function `1/sqrt(1+4x^2)`
Integrate the function `1/sqrt((2-x)^2 + 1)`
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(8+3x - x^2)`
Integrate the function `(4x+ 1)/sqrt(2x^2 + x - 3)`
Integrate the function `(x + 2)/sqrt(x^2 -1)`
Integrate the function `(x + 2)/sqrt(4x - x^2)`
Integrate the function `(x+2)/sqrt(x^2 + 2x + 3)`
Integrate the function `(x + 3)/(x^2 - 2x - 5)`
Integrate the function `(5x + 3)/sqrt(x^2 + 4x + 10)`
`int dx/sqrt(9x - 4x^2)` equals:
Integrate the function:
`sqrt(x^2 + 4x +1)`
Integrate the function:
`sqrt(1-4x - x^2)`
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 ______.
\[\int\frac{8x + 13}{\sqrt{4x + 7}} \text{ dx }\]
Find:
`int_(-pi/4)^0 (1+tan"x")/(1-tan"x") "dx"`
If θ f(x) = `int_0^x t sin t dt` then `f^1(x)` is
Find: `int (dx)/(x^2 - 6x + 13)`
`int (a^x - b^x)^2/(a^xb^x)dx` equals ______.
