Advertisements
Advertisements
प्रश्न
Find `int (sin^-1x)/(1 - x^2)^(3//2) dx`.
Advertisements
उत्तर
Let I = `int (sin^-1x)/(1 - x^2)^(3//2) dx`
Consider t = sin–1 x
`dt/dx = 1/sqrt(1 - x^2)`
∴ I = `int (t.dt)/((1 - x^2))`
= `int (t.dt)/((1 - sin^2t))`
= `int (t.dt)/(cos^2t)`
= `int t . sec^2 t dt`
On integrating by parts
= `t int sec^2t.dt - int {(d(t))/dt int sec^2 t}dt`
= `t tan t - int 1.tan t dt`
= t tan t – log sec t + C
= sin–1x tan [sin–1x] – log sec [sin–1x] + C
APPEARS IN
संबंधित प्रश्न
Prove that:
`int sqrt(x^2 - a^2)dx = x/2sqrt(x^2 - a^2) - a^2/2log|x + sqrt(x^2 - a^2)| + c`
Evaluate `int_0^(pi)e^2x.sin(pi/4+x)dx`
Integrate the function in x cos-1 x.
Integrate the function in `(x cos^(-1) x)/sqrt(1-x^2)`.
Integrate the function in ex (sinx + cosx).
Evaluate the following : `int log(logx)/x.dx`
Evaluate the following : `int cos(root(3)(x)).dx`
Integrate the following functions w.r.t. x: `sqrt(x^2 + 2x + 5)`.
Integrate the following functions w.r.t. x : `e^x/x [x (logx)^2 + 2 (logx)]`
Choose the correct options from the given alternatives :
`int tan(sin^-1 x)*dx` =
Choose the correct options from the given alternatives :
`int [sin (log x) + cos (log x)]*dx` =
Integrate the following with respect to the respective variable : cos 3x cos 2x cos x
Evaluate the following.
`int x^2 e^4x`dx
`int (sinx)/(1 + sin x) "d"x`
`int 1/sqrt(2x^2 - 5) "d"x`
`int (cos2x)/(sin^2x cos^2x) "d"x`
`int 1/(x^2 - "a"^2) "d"x` = ______ + c
`int log x * [log ("e"x)]^-2` dx = ?
`int "e"^x [x (log x)^2 + 2 log x] "dx"` = ______.
Evaluate the following:
`int_0^1 x log(1 + 2x) "d"x`
`int "dx"/(sin(x - "a")sin(x - "b"))` is equal to ______.
Solve: `int sqrt(4x^2 + 5)dx`
The integral `int x cos^-1 ((1 - x^2)/(1 + x^2))dx (x > 0)` is equal to ______.
Evaluate `int(1 + x + (x^2)/(2!))dx`
Evaluate:
`intcos^-1(sqrt(x))dx`
Evaluate:
`int e^(logcosx)dx`
Evaluate:
`int (logx)^2 dx`
If ∫(cot x – cosec2 x)ex dx = ex f(x) + c then f(x) will be ______.
Evaluate the following.
`intx^3/sqrt(1+x^4)`dx
