हिंदी

Choose the correct options from the given alternatives : ∫dxcosxsin2x-cos2x⋅dx =

Advertisements
Advertisements

प्रश्न

Choose the correct options from the given alternatives : 

`int dx/(cosxsqrt(sin^2x - cos^2x))*dx` =

विकल्प

  • `log(tanx - sqrt(tan^2x - 1)) + c`

  • sin–1 (tan x) + c

  • 1 + sin–1 (cot x) + c

  • `log(tanx + sqrt(tan^2x - 1)) + c`

MCQ
Advertisements

उत्तर

`log(tanx + sqrt(tan^2x - 1)) + c`

[ Hint : `int dx/(cosxsqrt(sin^2x - cos^2x)`

= `int (sec2x*dx)/sqrt(tan2x - 1)`     ...[Dividing by cos2x]
Put tan x = t].

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 1.16 | पृष्ठ १४९

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

Integrate the functions:

`(log x)^2/x`


Integrate the functions:

`x/(9 - 4x^2)`


Integrate the functions:

`(1+ log x)^2/x`


Evaluate : `∫1/(3+2sinx+cosx)dx`


Evaluate: `int_0^3 f(x)dx` where f(x) = `{(cos 2x, 0<= x <= pi/2),(3, pi/2 <= x <= 3) :}`


Evaluate: `int (sec x)/(1 + cosec x) dx`


Write a value of\[\int\frac{\sin x}{\cos^3 x} \text{ dx }\]


Write a value of\[\int\frac{\sin x - \cos x}{\sqrt{1 + \sin 2x}} \text{ dx}\]


\[\int x \sin^3 x\ dx\]

Evaluate the following integral: 

`int(4x + 3)/(2x + 1).dx`


Integrate the following functions w.r.t. x : `(sinx + 2cosx)/(3sinx + 4cosx)`


Integrate the following functions w.r.t. x : `(4e^x - 25)/(2e^x - 5)`


Evaluate the following : `int (1)/sqrt(11 - 4x^2).dx`


Evaluate the following : `int (1)/sqrt(x^2 + 8x - 20).dx`


Evaluate the following:

`int sinx/(sin 3x)  dx`


Evaluate the following integrals : `int (2x + 3)/(2x^2 + 3x - 1).dx`


Evaluate the following integrals:

`int (7x + 3)/sqrt(3 + 2x - x^2).dx`


Choose the correct options from the given alternatives :

`int f x^x (1 + log x)*dx`


Choose the correct options from the given alternatives :

`2 int (cos^2x - sin^2x)/(cos^2x + sin^2x)*dx` =


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


Evaluate the following.

`int "x" sqrt(1 + "x"^2)` dx


Evaluate the following.

`int "x"^3/(16"x"^8 - 25)` dx


Evaluate the following.

`int 1/("a"^2 - "b"^2 "x"^2)` dx


Evaluate the following.

`int 1/(sqrt(3"x"^2 + 8))` dx


Evaluate the following.

`int 1/(sqrt("x"^2 + 4"x"+ 29))` dx


Evaluate the following.

`int 1/(sqrt("x"^2 -8"x" - 20))` dx


`int 1/sqrt((x - 3)(x + 2))` dx = ______.


`int (2(cos^2 x - sin^2 x))/(cos^2 x + sin^2 x)` dx = ______________


`int ("e"^(2x) + "e"^(-2x))/("e"^x)  "d"x`


Evaluate  `int"e"^x (1/x - 1/x^2)  "d"x`


`int(5x + 2)/(3x - 4) dx` = ______


`int1/(4 + 3cos^2x)dx` = ______ 


`int "e"^(sin^-1 x) ((x + sqrt(1 - x^2))/(sqrt1 - x^2)) "dx" = ?`


`int (cos x)/(1 - sin x) "dx" =` ______.


`int sec^6 x tan x   "d"x` = ______.


`int ("d"x)/(x(x^4 + 1))` = ______.


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


Find `int dx/sqrt(sin^3x cos(x - α))`.


Evaluate.

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


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

`1/(x^2 +1) dx = tan ^-1 + c`


Evaluate the following

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


Evaluate.

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


Evaluate:

`int sin^3x cos^3x  dx`


Evaluate the following.

`intxsqrt(1+x^2)dx`


Evaluate the following

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


Evaluate `int 1/(x(x-1))dx`


Evaluate the following:

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


Which standard substitution is used for \[\sqrt{\mathrm{a}^2-x^2}\], \[\frac{1}{\sqrt{\mathrm{a}^2-x^2}}\], or \[\mathrm{a}^2-x^2\]?


What is \[\int\cot t\,dt\] in the evaluation of \[\int\frac{\sin x}{\sin(x+a)}\,dx\]?


Integration by substitution is the reverse process of which rule?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×