Advertisements
Advertisements
प्रश्न
Find `intsqrtx/sqrt(a^3-x^3)dx`
Advertisements
उत्तर
`I=intsqrtx/sqrt(a^3-x^3)dx`
Let: `x^(3/2)=t`
`=>3/2x^(1/2)dx=dt`
`x^(1/2)dx=2/3dt`
Putting the values in I, we get
`I=intsqrtx/sqrt(a^3-x^3)dx`
`=2/3int1/(sqrt(a^3-t^2))dt`
Using the following formula of integration, we get
`intdx/sqrt(a^2-x^2)=sin^(-1)(x/a)`
`:.2/3int1/sqrt(a^3-t^2)dt=2/3sin^(-1)(t/(a^(3/2)))+C`
Again, putting the value of t, we get
`2/3int1/sqrt(a^3-t^2)dt=2/3sin^(-1)(t/a^(3/2))+C`
`=2/3sin^(-1)(x^(3/2)/a^(3/2))+C`
Here, C is constant of integration.
APPEARS IN
संबंधित प्रश्न
Integrate the functions:
`x/(e^(x^2))`
Integrate the functions:
`(sin x)/(1+ cos x)^2`
`int (dx)/(sin^2 x cos^2 x)` equals:
Evaluate: `int 1/(x(x-1)) dx`
Evaluate the following integrals : `int (cos2x)/(sin^2x.cos^2x)dx`
Evaluate the following integrals : `int cos^2x.dx`
Integrate the following functions w.r.t. x : `(e^(2x) + 1)/(e^(2x) - 1)`
Evaluate the following : `int (1)/(1 + x - x^2).dx`
Evaluate the following:
`int sinx/(sin 3x) dx`
Integrate the following functions w.r.t. x : `int (1)/(2 + cosx - sinx).dx`
Integrate the following functions w.r.t. x : `int (1)/(3 - 2cos 2x).dx`
If f'(x) = x2 + 5 and f(0) = −1, then find the value of f(x).
Fill in the Blank.
`int (5("x"^6 + 1))/("x"^2 + 1)` dx = x4 + ______ x3 + 5x + c
`int e^x/x [x (log x)^2 + 2 log x]` dx = ______________
`int sqrt(x^2 + 2x + 5)` dx = ______________
`int cos sqrtx` dx = _____________
`int (sin4x)/(cos 2x) "d"x`
`int 1/(xsin^2(logx)) "d"x`
`int cos^7 x "d"x`
`int sin^-1 x`dx = ?
If f'(x) = `x + 1/x`, then f(x) is ______.
Evaluate `int_(logsqrt(2))^(logsqrt(3)) 1/((e^x + e^-x)(e^x - e^-x)) dx`.
Evaluate the following
`int1/(x^2 +4x-5)dx`
Evaluate:
`int 1/(1 + cosα . cosx)dx`
Evaluate `int (1+x+x^2/(2!)) dx`
Evaluate the following
`int x^3/sqrt(1+x^4) dx`
Evaluate `int (1 + x + x^2/(2!)) dx`
Evaluate `int(1+x+x^2/(2!))dx`
