हिंदी

Choose the correct options from the given alternatives : ∫1cosx-cos2x⋅dx =

Advertisements
Advertisements

प्रश्न

Choose the correct options from the given alternatives :

`int (1)/(cosx - cos^2x)*dx` =

विकल्प

  • `log ("cosec"x - cotx) + tan(x/2) + c`

  • sin 2x – cos x + c

  • `log (secx + tanx) - cot(x/2) + c`

  • cos 2x – sin x + c

MCQ
Advertisements

उत्तर

`log (secx + tanx) - cot(x/2) + c`

[ Hint : `int 1/(cosx - cos^2x)*dx`

= `int 1/(cosx(1 - cosx))*dx`

= `int ((1 - cosx) + cosx)/(cosx(1 - cosx))*dx`

= `int (1/cosx + 1/(1 - cosx))*dx`

= `int [sec x + 1/2 "cosec"^2(x/2)]*dx`

= `log|secx + tanx|1/2((-cotx/2))/(1/2) + c`

= `log|secx + tanx| - cot(x/2) + c`].

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Indefinite Integration - Miscellaneous Exercise 3 [पृष्ठ १४९]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Indefinite Integration
Miscellaneous Exercise 3 | Q 1.09 | पृष्ठ १४९

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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


If u and v are two functions of x then prove that

`intuvdx=uintvdx-int[du/dxintvdx]dx`

Hence evaluate, `int xe^xdx`


Integrate the function in x sin x.


Integrate the function in `((x- 3)e^x)/(x - 1)^3`.


Integrate the function in e2x sin x.


Integrate the function in `sin^(-1) ((2x)/(1+x^2))`.


`intx^2 e^(x^3) dx` equals: 


Evaluate the following : `int log(logx)/x.dx`


Evaluate the following : `int x.cos^3x.dx`


Evaluate the following : `int cos(root(3)(x)).dx`


Integrate the following functions w.r.t. x : `e^(2x).sin3x`


Integrate the following functions w.r.t. x : `sec^2x.sqrt(tan^2x + tan x - 7)`


Integrate the following functions w.r.t. x : `((1 + sin x)/(1 + cos x)).e^x`


Integrate the following functions w.r.t. x : `e^x/x [x (logx)^2 + 2 (logx)]`


Integrate the following functions w.r.t. x : `log(1 + x)^((1 + x)`


Choose the correct options from the given alternatives :

`int sin (log x)*dx` =


Integrate the following w.r.t.x : `log (1 + cosx) - xtan(x/2)`


Integrate the following w.r.t.x : log (log x)+(log x)–2 


Evaluate the following.

`int (log "x")/(1 + log "x")^2` dx


Evaluate: `int ("ae"^("x") + "be"^(-"x"))/("ae"^("x") - "be"^(−"x"))` dx


Evaluate: `int "dx"/sqrt(4"x"^2 - 5)`


Evaluate: `int e^x/sqrt(e^(2x) + 4e^x + 13)` dx


`int "e"^x x/(x + 1)^2  "d"x`


`int logx/(1 + logx)^2  "d"x`


Evaluate the following:

`int (sin^-1 x)/((1 - x)^(3/2)) "d"x`


Evaluate the following:

`int ((cos 5x + cos 4x))/(1 - 2 cos 3x) "d"x`


Evaluate the following:

`int_0^1 x log(1 + 2x)  "d"x`


`int "dx"/(sin(x - "a")sin(x - "b"))` is equal to ______.


`int tan^-1 sqrt(x)  "d"x` is equal to ______.


Find: `int (2x)/((x^2 + 1)(x^2 + 2)) dx`


Find the general solution of the differential equation: `e^((dy)/(dx)) = x^2`.


`int 1/sqrt(x^2 - a^2)dx` = ______.


`int((4e^x - 25)/(2e^x - 5))dx = Ax + B log(2e^x - 5) + c`, then ______.


`int(1-x)^-2 dx` = ______


`int1/sqrt(x^2 - a^2) dx` = ______


Evaluate the following.

`int x^3 e^(x^2) 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 the following.

`intx^3/sqrt(1+x^4)dx`


Evaluate the following.

`intx^3/sqrt(1+x^4)  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`  


Evaluate the following.

`intx^3/sqrt(1+x^4)`dx


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


The value of `int (x sin^-1)/(sqrt(1 - x^2)) dx` is equal to:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×