Advertisements
Advertisements
प्रश्न
Integrate the following functions with respect to x:
`sqrt(9 - (2x + 5)^2`
Advertisements
उत्तर
`int sqrt(9 - (2x + 5)^2) "d"x = int sqrt(3^2 - (2x + 5)^2) "d"x`
Put 2x + 5 = t
2 dx = dt
dx = `1/2 "dt"`
= `int sqrt(3^2 - "t"^2) xx 1/2 "dt"`
= `1/2 int sqrt(3^2 - "t"^2) "dt"`
= `1/2 ["t"/2 sqrt(3^2 - "t"^2) + 3^2/2 sin^-1 ("t"/3) + "c"]`
= `1/4 [(2x + 5) sqrt(9 - (2x + 5)^2) + 9 sin^-1 ((2x + 5)/3)] + "c"`
`int sqrt(9 - (2x + 5)^2) "d"x = 1/4 [(2x + 5) sqrt(9 - (2x + 5)^2) + 9 sin^-1 ((2x + 5)/3)] + "c"`
APPEARS IN
संबंधित प्रश्न
Evaluate : `int1/(x(3+logx))dx`
Integrate the following functions with respect to x :
`(cos2x - cos 2 alpha)/(cosx - cos alpha)`
Integrate the following functions with respect to x :
`1/(sqrt(x + 3) - sqrt(x - 4))`
Integrate the following with respect to x :
`x^2/(1 + x^6)`
Integrate the following with respect to x :
`("e"^x - "e"^-x)/("e"^x + "e"^-x)`
Integrate the following with respect to x:
27x2e3x
Integrate the following with respect to x:
`tan^-1 ((8x)/(1 - 16x^2))`
Integrate the following with respect to x:
`"e"^(- 3x) cos x`
Integrate the following with respect to x:
`"e"^x (tan x + log sec x)`
Integrate the following with respect to x:
`"e"^x ((x - 1)/(2x^2))`
Find the integrals of the following:
`1/((x + 1)^2 - 25)`
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 sqrt(tanx)/(sin2x) "d"x` is
Choose the correct alternative:
`int ("e"^(6 log x) - "e"^(5logx))/("e"^(4logx) - "e"^(3logx)) "d"x` is
Choose the correct alternative:
`int secx/sqrt(cos2x) "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 (sec^2x)/(tan^2 x - 1) "d"x`
Choose the correct alternative:
`int sin sqrt(x) "d"x` is
