Advertisements
Advertisements
प्रश्न
`int "dx"/(sin(x - "a")sin(x - "b"))` is equal to ______.
पर्याय
`sin("b" - "a") log|(sin(x - "b"))/(sin(x - "a"))| + "C"`
`"cosec"("b" - "a") log|(sin(x - "a"))/(sin(x - "b"))| + "C"`
`"cosec"("b" - "a") log|(sin(x - "b"))/(sin(x - "a"))| + "C"`
`sin("b" - "a")log|(sin("x" - "a"))/(sin(x - "b"))| + "C"`
Advertisements
उत्तर
`int "dx"/(sin(x - "a")sin(x - "b"))` is equal to `"cosec"("b" - "a") log|(sin(x - "b"))/(sin(x - "a"))| + "C"`.
Explanation:
Let I = `int "dx"/(sin(x - "a")sin(x - "b"))`
Multiplying and dividing by sin(b – a) we get,
I = `1/(sin("b" - "a")) int (sin("b" - "a"))/(sin(x - "a") * sin(x - "b")) "d"x`
= `1/(sin("b" - "a")) int (sin(x + "b" - x - "a"))/(sin(x - "a") * sin(x - "b")) "d"x`
= `1/(sin("b" - "a")) int (sin[(x - "a") - (x - "b")])/(sin(x - "a") * sin(x - "b")) "d"x`
= `1/(sin("b" - "a")) int (sin(x - "a") cos(x - "b") - cos(x - "a") sin(x - "b"))/(sin(x - "a") * sin(x - "b")) "d"x`
= `1/(sin("b" - "a")) int (sin(x - "a") * cos(x - "b"))/(sin(x - "a")*sin(x - "b")) - (cos(x - "a")*sin(x - "b"))/(sin(x - "a") * sin(x - "b")) "d"x`
= `1/(sin("b" - "a")) int [(cos(x - "b"))/(sin(x - "b")) - (cos(x - "a"))/(sin(x - "a"))]"d"x`
= `1/(sin("b" - "a")) int [cot(x - "b") - cot(x - "a")]"d"x`
= `1/(sin("b" - "a")) [log sin(x - "b") - logsin(x - "a")] + "C"`
= `1/(sin("b" - "a")) * log|(sin(x - "b"))/(sin(x - "a"))| + "C"`
I = `"cosec"("b" - "a") log|(sin(x - "b"))/(sin(x - "a"))| + "C"`.
APPEARS IN
संबंधित प्रश्न
Integrate the function in x sin x.
Integrate the function in (sin-1x)2.
Integrate the function in x sec2 x.
Integrate the function in `e^x (1/x - 1/x^2)`.
Integrate the function in `((x- 3)e^x)/(x - 1)^3`.
`int e^x sec x (1 + tan x) dx` equals:
Evaluate the following : `int x^2.log x.dx`
Evaluate the following:
`int x^2 sin 3x dx`
Evaluate the following : `int x^2*cos^-1 x*dx`
Evaluate the following : `int cos(root(3)(x)).dx`
Integrate the following functions w.r.t. x:
sin (log x)
Integrate the following functions w.r.t. x : `sqrt(2x^2 + 3x + 4)`
Choose the correct options from the given alternatives :
`int (1)/(cosx - cos^2x)*dx` =
Integrate the following with respect to the respective variable : `t^3/(t + 1)^2`
Integrate the following with respect to the respective variable : `(3 - 2sinx)/(cos^2x)`
Integrate the following w.r.t.x : `(1)/(xsin^2(logx)`
Integrate the following w.r.t.x : e2x sin x cos x
Evaluate the following.
`int "e"^"x" [(log "x")^2 + (2 log "x")/"x"]` dx
Evaluate: Find the primitive of `1/(1 + "e"^"x")`
Evaluate: `int "dx"/(25"x" - "x"(log "x")^2)`
`int (sin(x - "a"))/(cos (x + "b")) "d"x`
`int sin4x cos3x "d"x`
`int"e"^(4x - 3) "d"x` = ______ + c
`int (x^2 + x - 6)/((x - 2)(x - 1)) "d"x` = x + ______ + c
Evaluate `int 1/(x(x - 1)) "d"x`
Evaluate `int 1/(x log x) "d"x`
`int_0^"a" sqrt("x"/("a" - "x")) "dx"` = ____________.
`int "e"^x int [(2 - sin 2x)/(1 - cos 2x)]`dx = ______.
The integral `int x cos^-1 ((1 - x^2)/(1 + x^2))dx (x > 0)` is equal to ______.
If `int (f(x))/(log(sin x))dx` = log[log sin x] + c, then f(x) is equal to ______.
Find `int e^x ((1 - sinx)/(1 - cosx))dx`.
Solve the differential equation (x2 + y2) dx - 2xy dy = 0 by completing the following activity.
Solution: (x2 + y2) dx - 2xy dy = 0
∴ `dy/dx=(x^2+y^2)/(2xy)` ...(1)
Puty = vx
∴ `dy/dx=square`
∴ equation (1) becomes
`x(dv)/dx = square`
∴ `square dv = dx/x`
On integrating, we get
`int(2v)/(1-v^2) dv =intdx/x`
∴ `-log|1-v^2|=log|x|+c_1`
∴ `log|x| + log|1-v^2|=logc ...["where" - c_1 = log c]`
∴ x(1 - v2) = c
By putting the value of v, the general solution of the D.E. is `square`= cx
Evaluate:
`intcos^-1(sqrt(x))dx`
If f′(x) = 4x3 − 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x).
Evaluate:
`int x^2 cos x dx`
Evaluate the following.
`int x sqrt(1 + x^2) dx`
The value of `inta^x.e^x dx` equals
