Advertisements
Advertisements
प्रश्न
Find the integrals of the function:
`(sin^2 x)/(1 + cos x)`
Advertisements
उत्तर
Let `I = int (sin^2 x)/(1 + cos x) dx`
= `int (1 - cos^2 x)/(1+ cos x) dx`
= `int ((1 - cos x) (1 + cos x))/(1 + cos x) dx`
= `int (1 - cos x) dx`
= `int 1 dx - int cos x dx`
= x - sin x + C
APPEARS IN
संबंधित प्रश्न
Evaluate :`int_(pi/6)^(pi/3) dx/(1+sqrtcotx)`
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:
sin3 x cos3 x
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:
cos4 2x
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:
`(cos 2x+ 2sin^2x)/(cos^2 x)`
Find `int dx/(x^2 + 4x + 8)`
Differentiate : \[\tan^{- 1} \left( \frac{1 + \cos x}{\sin x} \right)\] with respect to x .
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_ (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_ (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`
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 (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 sinx/(3 + 4cos^2x) "d"x` = ______.
The value of the integral `int_(1/3)^1 (x - x^3)^(1/3)/x^4 dx` is
