मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Choose the correct options from the given alternatives : ∫sin(logx)⋅dx = - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Choose the correct options from the given alternatives :

`int sin (log x)*dx` =

पर्याय

  • `x/(2)[sin (log x) - cos (log x)] + c`

  • `x/(2)[sin (log x) + cos (log x)] + c`

  • `x/(2)[cos (log x) - sin (log x)] + c`

  • `x/(4)[cos (log x) - sin (log x)] + c`

MCQ
Advertisements

उत्तर

`x/(2)[sin (log x) - cos (log x)] + c`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Indefinite Integration - Miscellaneous Exercise 3 [पृष्ठ १४९]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 3 Indefinite Integration
Miscellaneous Exercise 3 | Q 1.12 | पृष्ठ १४९

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

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`


If `int_(-pi/2)^(pi/2)sin^4x/(sin^4x+cos^4x)dx`, then the value of I is:

(A) 0

(B) π

(C) π/2

(D) π/4


Integrate the function in `(x cos^(-1) x)/sqrt(1-x^2)`.


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^2.log x.dx`


Integrate the following functions w.r.t. x : `sqrt(5x^2 + 3)`


Integrate the following functions w.r.t. x : `x^2 .sqrt(a^2 - x^6)`


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^(5x).[(5x.logx + 1)/x]`


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 : `(sin^6θ + cos^6θ)/(sin^2θ*cos^2θ)`


Evaluate the following.

`int x^2 *e^(3x)`dx


Evaluate the following.

`int "e"^"x" "x - 1"/("x + 1")^3` dx


`int ("x" + 1/"x")^3 "dx"` = ______


Choose the correct alternative from the following.

`int (1 - "x")^(-2) "dx"` = 


Evaluate: `int "e"^"x"/(4"e"^"2x" -1)` dx


`int (sin(x - "a"))/(cos (x + "b"))  "d"x`


`int ("e"^xlog(sin"e"^x))/(tan"e"^x)  "d"x`


`int sqrt(tanx) + sqrt(cotx)  "d"x`


Choose the correct alternative:

`intx^(2)3^(x^3) "d"x` =


`int ("d"x)/(x - x^2)` = ______


`int(x + 1/x)^3 dx` = ______.


`int 1/x  "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/(4x^2 - 1)  "d"x`


`int logx/(1 + logx)^2  "d"x`


`int "e"^x [x (log x)^2 + 2 log x] "dx"` = ______.


`int "e"^x int [(2 - sin 2x)/(1 - cos 2x)]`dx = ______.


Find `int_0^1 x(tan^-1x)  "d"x`


Evaluate the following:

`int_0^pi x log sin x "d"x`


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`


The integral `int x cos^-1 ((1 - x^2)/(1 + x^2))dx (x > 0)` is equal to ______.


`int_0^1 x tan^-1 x  dx` = ______.


Find `int e^x ((1 - sinx)/(1 - cosx))dx`.


Find: `int e^(x^2) (x^5 + 2x^3)dx`.


Solution of the equation `xdy/dx=y log y` is ______


Evaluate `int(3x-2)/((x+1)^2(x+3))  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


`int logx  dx = x(1+logx)+c`


Solve the following

`int_0^1 e^(x^2) x^3 dx`


Evaluate:

`int((1 + sinx)/(1 + cosx))e^x dx`


Evaluate:

`int (logx)^2 dx`


Evaluate the following.

`intx^3  e^(x^2) dx`


Evaluate the following.

`intx^3/sqrt(1+x^4)  dx`


The value of `inta^x.e^x dx` equals


Evaluate `int(1 + x + x^2/(2!))dx`.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×