Advertisements
Advertisements
प्रश्न
Integrate the following functions w.r.t. x : cosec (log x)[1 – cot (log x)]
Advertisements
उत्तर
Let I = `int "cosec" (log x)[1 - cot (log x)].dx`
Put log x = t
∴ et
∴ dx = et .dt
∴ I = `int "cosec" t (1 - cot t).e^t dt`
= `int e^t ["cosec" t - "cosec" t cot t].dt`
= `int e^t ["cosec" t + d/dt ("cosec" t)].dt`
= `e^t "cosec" t + c ...[∵ int e^t [f(t) + f'(t)].dt = e^t f(t) + c]`
= x . cosec (log x) + c.
APPEARS IN
संबंधित प्रश्न
Prove that: `int sqrt(a^2 - x^2) * dx = x/2 * sqrt(a^2 - x^2) + a^2/2 * sin^-1(x/a) + c`
`int1/xlogxdx=...............`
(A)log(log x)+ c
(B) 1/2 (logx )2+c
(C) 2log x + c
(D) log x + c
Integrate the function in x (log x)2.
Integrate the function in (x2 + 1) log x.
Integrate the function in `e^x (1 + sin x)/(1+cos x)`.
Integrate the function in e2x sin x.
Prove that:
`int sqrt(x^2 + a^2)dx = x/2 sqrt(x^2 + a^2) + a^2/2 log |x + sqrt(x^2 + a^2)| + c`
Evaluate the following : `int x.sin^2x.dx`
Evaluate the following: `int logx/x.dx`
Evaluate the following : `int cos(root(3)(x)).dx`
Integrate the following functions w.r.t. x : `sqrt(5x^2 + 3)`
Integrate the following functions w.r.t. x : `sec^2x.sqrt(tan^2x + tan x - 7)`
Integrate the following functions w.r.t. x : `e^(sin^-1x)*[(x + sqrt(1 - x^2))/sqrt(1 - x^2)]`
Choose the correct options from the given alternatives :
`int (sin^m x)/(cos^(m+2)x)*dx` =
Choose the correct options from the given alternatives :
`int [sin (log x) + cos (log x)]*dx` =
Integrate the following w.r.t. x: `(1 + log x)^2/x`
Integrate the following w.r.t.x : log (log x)+(log x)–2
Integrate the following w.r.t.x : `(1)/(x^3 sqrt(x^2 - 1)`
Evaluate the following.
`int e^x (1/x - 1/x^2)`dx
`int ("x" + 1/"x")^3 "dx"` = ______
Choose the correct alternative from the following.
`int (("x"^3 + 3"x"^2 + 3"x" + 1))/("x + 1")^5 "dx"` =
Evaluate: `int e^x/sqrt(e^(2x) + 4e^x + 13)` dx
Evaluate: `int "dx"/(5 - 16"x"^2)`
`int (sin(x - "a"))/(cos (x + "b")) "d"x`
`int 1/sqrt(2x^2 - 5) "d"x`
`int (cos2x)/(sin^2x cos^2x) "d"x`
Evaluate `int 1/(x log x) "d"x`
`int 1/sqrt(x^2 - 8x - 20) "d"x`
`int_0^"a" sqrt("x"/("a" - "x")) "dx"` = ____________.
`int cot "x".log [log (sin "x")] "dx"` = ____________.
`int log x * [log ("e"x)]^-2` dx = ?
`int "e"^x [x (log x)^2 + 2 log x] "dx"` = ______.
Evaluate the following:
`int (sin^-1 x)/((1 - x)^(3/2)) "d"x`
`int tan^-1 sqrt(x) "d"x` is equal to ______.
If u and v ore differentiable functions of x. then prove that:
`int uv dx = u intv dx - int [(du)/(d) intv dx]dx`
Hence evaluate `intlog x dx`
Evaluate: `int_0^(pi/4) (dx)/(1 + tanx)`
The integral `int x cos^-1 ((1 - x^2)/(1 + x^2))dx (x > 0)` is equal to ______.
`int(1-x)^-2 dx` = ______
`intsqrt(1+x) dx` = ______
Solution of the equation `xdy/dx=y log y` is ______
Evaluate:
`int(1+logx)/(x(3+logx)(2+3logx)) 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 `int tan^-1x dx`
If u and v are two differentiable functions of x, then prove that `intu*v*dx = u*intv dx - int(d/dx u)(intv dx)dx`. Hence evaluate: `intx cos x dx`
Evaluate the following.
`intx^3/sqrt(1+x^4)dx`
Evaluate:
`int x^2 cos x dx`
