Advertisements
Advertisements
Question
Find the integrals of the function:
`cos x/(1 + cos x)`
Advertisements
Solution
Let `I = int cos x/(1 + cos x) dx`
`= int ((1 + cos x) - 1)/(1 + cos x) dx`
`= int 1 dx - int 1/ (1 + cos x) dx`
`= int dx - int 1/ (2 cos^2 x/2) dx`
`= int dx - 1/2 int sec^2 x/2 dx`
`= x = 1/2* (tan x/2)/(1/2) + C`
`= x - tan x/2 + C`
APPEARS IN
RELATED QUESTIONS
Evaluate : `intsin(x-a)/sin(x+a)dx`
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:
cos4 2x
Find the integrals of the function:
`(sin^2 x)/(1 + cos x)`
Find the integrals of the function:
`(cos 2x - cos 2 alpha)/(cos x - cos alpha)`
Find the integrals of the function:
tan3 2x sec 2x
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`
Evaluate `int_0^(3/2) |x sin pix|dx`
Evaluate : \[\int\limits_0^\pi \frac{x \tan x}{\sec x \cdot cosec x}dx\] .
Find `int_ sin ("x" - a)/(sin ("x" + a )) d"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.
Integrate the function `cos("x + a")/sin("x + b")` w.r.t. x.
Find: `int sec^2 x /sqrt(tan^2 x+4) dx.`
Find: `int sin^-1 (2x) dx.`
Find `int "dx"/(2sin^2x + 5cos^2x)`
Find `int x^2tan^-1x"d"x`
`int "dx"/(sin^2x cos^2x)` is equal to ______.
`int (sin^6x)/(cos^8x) "d"x` = ______.
Evaluate the following:
`int ((1 + cosx))/(x + sinx) "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 sin^-1 sqrt(x/("a" + x)) "d"x` (Hint: Put x = a tan2θ)
`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
