Advertisements
Advertisements
Question
Evaluate the following:
`int ("d"x)/(1 + cos x)`
Solution
I = `int ("d"x)/(1 + cos x)`
= `int 1/(2 cos^2 x/2) "d"x`
= `1/2 int sec^2 x/2 "d"x`
= `1/2 * 1/(1/2) tan x/2 + "C"`
= `tan x/2 + "C"`
APPEARS IN
RELATED QUESTIONS
Evaluate : `intsin(x-a)/sin(x+a)dx`
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 x sin 2x sin 3x
Find the integrals of the function:
`cos x/(1 + cos x)`
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:
`(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:
sin−1 (cos x)
Find `int (sin^2 x - cos^2x)/(sin x cos x) dx`
Find `int((3 sin x - 2) cos x)/(13 - cos^2 x- 7 sin x) dx`
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_ 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.
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 sec^2 x /sqrt(tan^2 x+4) dx.`
Evaluate `int tan^8 x sec^4 x"d"x`
`int "e"^x (cosx - sinx)"d"x` is equal to ______.
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^6x + cos^6x)/(sin^2x cos^2x) "d"x`
Evaluate the following:
`int (cosx - cos2x)/(1 - cosx) "d"x`
