Advertisements
Advertisements
प्रश्न
Find: `int sec^2 x /sqrt(tan^2 x+4) dx.`
Advertisements
उत्तर
`I = int sec^2 x /sqrt(tan^2 x+4) dx.`
`"Let" tan x = t`
`sec^2 xdx = dt`
So, `I = int dt/sqrt(t^2 + 4)`
or, `I = int dt/sqrt(t^2 + 2^2)`
Since, we Know
`int dx/sqrt(x^2 + a^2) = "In" |x + sqrt(a^2 + x^2)| + "C"`
`I = "In" |t + sqrt(t^2 + 4)| + "C"`
`i.e`
`I = "In"|tanx + sqrt(tan^2 x + 4)| + "C"`
APPEARS IN
संबंधित प्रश्न
Find the integrals of the function:
sin2 (2x + 5)
Find the integrals of the function:
sin 3x cos 4x
Find the integrals of the function:
sin3 (2x + 1)
Find the integrals of the function:
`(1-cosx)/(1 + cos x)`
Find the integrals of the function:
`cos x/(1 + cos x)`
Find the integrals of the function:
sin4 x
Find the integrals of the function:
cos4 2x
Find the integrals of the function:
`(sin^2 x)/(1 + cos x)`
Find the integrals of the function:
`1/(sin xcos^3 x)`
`int (sin^2x - cos^2 x)/(sin^2 x cos^2 x) dx` is equal to ______.
Evaluate : \[\int\limits_0^\pi \frac{x \tan x}{\sec x \cdot cosec x}dx\] .
Find `int_ (sin "x" - cos "x" )/sqrt(1 + sin 2"x") d"x", 0 < "x" < π / 2 `
Find `int_ (log "x")^2 d"x"`
Find: `int_ (cos"x")/((1 + sin "x") (2+ sin"x")) "dx"`
Find:
`int"dx"/sqrt(5-4"x" - 2"x"^2)`
Integrate the function `cos("x + a")/sin("x + b")` w.r.t. x.
Find: `int sin^-1 (2x) dx.`
Find `int x^2tan^-1x"d"x`
`int "e"^x (cosx - sinx)"d"x` is equal to ______.
Evaluate the following:
`int ((1 + cosx))/(x + sinx) "d"x`
Evaluate the following:
`int ("d"x)/(1 + cos x)`
Evaluate the following:
`int tan^2x sec^4 x"d"x`
Evaluate the following:
`int sqrt(1 + sinx)"d"x`
Evaluate the following:
`int (sin^6x + cos^6x)/(sin^2x cos^2x) "d"x`
`int (x + sinx)/(1 + cosx) "d"x` is equal to ______.
`int sinx/(3 + 4cos^2x) "d"x` = ______.
The value of the integral `int_(1/3)^1 (x - x^3)^(1/3)/x^4 dx` is
