Advertisements
Advertisements
प्रश्न
Choose the correct alternative:
`int sin sqrt(x) "d"x` is
विकल्प
`2(- sqrt(x) cos sqrt(x) + sin sqrt(x)) + "c"`
`2(- sqrt(x) cos sqrt(x) - sin sqrt(x)) + "c"`
`2(- sqrt(x) sin sqrt(x) - cos sqrt(x)) + "c"`
`2(- sqrt(x) sin sqrt(x) + cos sqrt(x)) + "c"`
Advertisements
उत्तर
`2(- sqrt(x) cos sqrt(x) + sin sqrt(x)) + "c"`
APPEARS IN
संबंधित प्रश्न
Evaluate : `int _0^1 ("x" . ("sin"^-1 "x")^2)/sqrt (1 - "x"^2)` dx
Integrate the following functions with respect to x :
`(cos2x - cos 2 alpha)/(cosx - cos alpha)`
Integrate the following functions with respect to x :
sin2 5x
Integrate the following with respect to x :
`("cosec" x)/(log(tan x/2))`
Integrate the following with respect to x :
`alpha beta x^(alpha - 1) "e"^(- beta x^alpha)`
Integrate the following with respect to x:
9xe3x
Integrate the following with respect to x:
25xe–5x
Integrate the following with respect to x:
x2 cos x
Integrate the following with respect to x:
x5ex2
Integrate the following with respect to x:
`"e"^x (tan x + log sec x)`
Integrate the following with respect to x:
`(x + 2)/sqrt(x^2 - 1)`
Integrate the following functions with respect to x:
`sqrt(9 - (2x + 5)^2`
Choose the correct alternative:
The gradient (slope) of a curve at any point (x, y) is `(x^2 - 4)/x^2`. If the curve passes through the point (2, 7), then the equation of the curve is
Choose the correct alternative:
`int ("e"^x (1 + x))/(cos^2(x"e"^x)) "d"x` is
Choose the correct alternative:
`int sin^2x "d"x` is
Choose the correct alternative:
`int tan^-1 sqrt((1 - cos 2x)/(1 + cos 2x)) "d"x` is
Choose the correct alternative:
`int sqrt((1 - x)/(1 + x)) "d"x` is
Choose the correct alternative:
`int (x + 2)/sqrt(x^2 - 1) "d"x` is
Choose the correct alternative:
`int 1/(x sqrt(log x)^2 - 5) "d"x` is
