Advertisements
Advertisements
प्रश्न
Integrate the following with respect to x:
x2 cos x
Advertisements
उत्तर
`int x^2 cos x "d"x`
u = x2
u' = 2x
u' = 2
u"' 0
dv = cos x dx
⇒ v= `int cos x "d"x`
= sin x
v1 = `int "v" "d"x`
= `int sinx "d"x`
= `- cos x`
v2 = `int "v"_1 "d"x`
= `int - cosx "d"x`
= `- sin x`
v3 = `int "v"_2 "d"x`
= `int- sin x "d"x`
= `- int sinx "d"x`
= `- (cos x)`
`int "u" "dv"` = uv – u'v1 + uv2 – u"'v3 + ...........
`int x^2 cos x "d"x = x^2 sin x - 2x xx cos x + 2 xx - sin x - 0 xx cos x + "c"`
`int x^2 cos x "d"x = x^2 sin x + 2x cosx - 2 sinx + "c"`
APPEARS IN
संबंधित प्रश्न
Evaluate : `int1/(x(3+logx))dx`
Evaluate : `∫_0^(pi/2) (sinx.cosx)/(1 + sin^4x)`.dx
Evaluate : `int _0^1 ("x" . ("sin"^-1 "x")^2)/sqrt (1 - "x"^2)` dx
Integrate the following functions with respect to x :
`(x^3 + 4x^2 - 3x + 2)/x^2`
Integrate the following functions with respect to x :
`1/((x - 1)(x + 2)^2`
Integrate the following with respect to x :
`alpha beta x^(alpha - 1) "e"^(- beta x^alpha)`
Integrate the following with respect to x:
x log x
Integrate the following with respect to x:
27x2e3x
Integrate the following with respect to x:
`sin^-1 ((2x)/(1 + x^2))`
Integrate the following with respect to x:\
`logx/(1 + log)^2`
Integrate the following with respect to x:
`(2x + 1)/sqrt(9 + 4x - x^2)`
Integrate the following with respect to x:
`(x + 2)/sqrt(x^2 - 1)`
Integrate the following functions with respect to x:
`sqrt((6 - x)(x - 4))`
Choose the correct alternative:
If `int 3^(1/x)/x^2 "d"x = "k"(3^(1/x)) + "c"`, then the value of k is
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(x^2 tan^-1x + tan^-1x + 1))/(x^2 + 1) "d"x` is
Choose the correct alternative:
`int ("d"x)/("e"^x - 1)` is
Choose the correct alternative:
`int 1/(x sqrt(log x)^2 - 5) "d"x` is
