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


Integrate the function in x log 2x.


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


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


Evaluate the following : `int x^2.log x.dx`


Evaluate the following:

`int x^2 sin 3x  dx`


Evaluate the following : `int x.sin^2x.dx`


Evaluate the following: `int logx/x.dx`


Integrate the following functions w.r.t.x:

`e^-x cos2x`


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 (1)/(x + x^5)*dx` = f(x) + c, then `int x^4/(x + x^5)*dx` =


Choose the correct options from the given alternatives :

`int sin (log x)*dx` =


Integrate the following w.r.t.x : log (x2 + 1)


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


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


Evaluate: `int "dx"/(3 - 2"x" - "x"^2)`


Evaluate: `int "dx"/("x"[(log "x")^2 + 4 log "x" - 1])`


Evaluate: `int "dx"/(25"x" - "x"(log "x")^2)`


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


`int (cos2x)/(sin^2x cos^2x)  "d"x`


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


Choose the correct alternative:

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


Evaluate `int 1/(x log x)  "d"x`


`int [(log x - 1)/(1 + (log x)^2)]^2`dx = ?


∫ log x · (log x + 2) dx = ?


Evaluate the following:

`int_0^1 x log(1 + 2x)  "d"x`


`int 1/sqrt(x^2 - 9) dx` = ______.


`int x/((x + 2)(x + 3)) dx` = ______ + `int 3/(x + 3) dx`


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


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


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


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


Evaluate :

`int(4x - 6)/(x^2 - 3x + 5)^(3/2)  dx`


`int(3x^2)/sqrt(1+x^3) dx = sqrt(1+x^3)+c`


Evaluate `int(3x-2)/((x+1)^2(x+3))  dx`


`int(f'(x))/sqrt(f(x)) dx = 2sqrt(f(x))+c`


Evaluate the following.

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


Evaluate:

`int (logx)^2 dx`


Evaluate `int tan^-1x  dx`


Evaluate:

`int (sin(x - a))/(sin(x + a))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  e^(x^2) dx`


Evaluate:

`int1/(x^2 + 25)dx`


Evaluate the following.

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


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×