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:
cos 2x cos 4x cos 6x
Find the integrals of the function:
`(sin^2 x)/(1 + cos x)`
Find the integrals of the function:
`(cos x - sinx)/(1+sin 2x)`
Find the integrals of the function:
tan3 2x sec 2x
Find the integrals of the function:
`(cos 2x+ 2sin^2x)/(cos^2 x)`
Find the integrals of the function:
`1/(sin xcos^3 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 ______.
Find `int (sin^2 x - cos^2x)/(sin x cos x) dx`
Find `int dx/(x^2 + 4x + 8)`
Evaluate `int_0^pi (x sin x)/(1 + cos^2 x) dx`
Evaluate `int_0^(3/2) |x sin pix|dx`
Find `int (2x)/((x^2 + 1)(x^4 + 4))`dx
Find `int_ (sin "x" - cos "x" )/sqrt(1 + sin 2"x") d"x", 0 < "x" < π / 2 `
Find `int_ sin ("x" - a)/(sin ("x" + a )) d"x"`
Find `int_ (log "x")^2 d"x"`
Find the area of the triangle whose vertices are (-1, 1), (0, 5) and (3, 2), using integration.
Find: `int_ (cos"x")/((1 + sin "x") (2+ sin"x")) "dx"`
Find: `int sec^2 x /sqrt(tan^2 x+4) dx.`
Find: `intsqrt(1 - sin 2x) dx, pi/4 < x < pi/2`
Find: `int sin^-1 (2x) dx.`
Evaluate `int tan^8 x sec^4 x"d"x`
Find `int x^2tan^-1x"d"x`
`int "e"^x (cosx - sinx)"d"x` is equal to ______.
`int (sin^6x)/(cos^8x) "d"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 (cosx - cos2x)/(1 - cosx) "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 ______.
`int (cos^2x)/(sin x + cos x)^2 dx` is equal to
