Advertisements
Advertisements
प्रश्न
Find the integrals of the function:
`(1-cosx)/(1 + cos x)`
उत्तर
Let `I = int (1 - cos x)/(1 + cos x) dx`
`= int (2 sin^2 x/2)/(2 cos^2 x/2) dx`
`= int tan^2 x/2 dx`
`= int (sec^2 x/2 - 1) dx`
`= [(tan x/2)/(1/2) - x + C]`
`= 2 tan x/2 - x + 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:
cos 2x cos 4x cos 6x
Find the integrals of the function:
sin3 x cos3 x
Find the integrals of the function:
sin 4x sin 8x
Find the integrals of the function:
sin4 x
Find the integrals of the function:
`(sin^3 x + cos^3 x)/(sin^2x cos^2 x)`
Find the integrals of the function:
`(cos 2x+ 2sin^2x)/(cos^2 x)`
Find the integrals of the function:
`(cos 2x)/(cos x + sin x)^2`
`int (e^x(1 +x))/cos^2(e^x x) dx` equals ______.
Evaluate `int_0^(3/2) |x sin pix|dx`
Find `int((3 sin x - 2) cos x)/(13 - cos^2 x- 7 sin x) dx`
Differentiate : \[\tan^{- 1} \left( \frac{1 + \cos x}{\sin x} \right)\] with respect to x .
Find `int_ (sin2"x")/((sin^2 "x"+1)(sin^2"x"+3))d"x"`
Find the area of the triangle whose vertices are (-1, 1), (0, 5) and (3, 2), using integration.
Find:
`int"dx"/sqrt(5-4"x" - 2"x"^2)`
Find: `int sec^2 x /sqrt(tan^2 x+4) dx.`
Find: `int sin^-1 (2x) dx.`
Find `int "dx"/(2sin^2x + 5cos^2x)`
`int "e"^x (cosx - sinx)"d"x` is equal to ______.
`int (sin^6x)/(cos^8x) "d"x` = ______.
Evaluate the following:
`int ((1 + cosx))/(x + sinx) "d"x`
Evaluate the following:
`int ("d"x)/(1 + cos x)`
Evaluate the following:
`int (sinx + cosx)/sqrt(1 + sin 2x) "d"x`
Evaluate the following:
`int sqrt(1 + sinx)"d"x`
Evaluate the following:
`int sin^-1 sqrt(x/("a" + x)) "d"x` (Hint: Put x = a tan2θ)
`int (x + sinx)/(1 + cosx) "d"x` is equal to ______.
The value of the integral `int_(1/3)^1 (x - x^3)^(1/3)/x^4 dx` is
`int (cos^2x)/(sin x + cos x)^2 dx` is equal to