हिंदी

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

Advertisements
Advertisements

प्रश्न

Choose the correct options from the given alternatives :

`int (cos2x - 1)/(cos2x + 1)*dx` =

विकल्प

  • tan x – x + c

  • x + tan x + c

  • x – tan x + c

  • – x – cot x + c

MCQ
Advertisements

उत्तर

x – tan x + c

[ Hint : `int (cos2x - 1)/(cos2x + 1)*dx`

= `int (-(1 - cos2x))/(1 + cos^2x)*dx`

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

= `int (sec^2x - 1)*dx`

= – tan x + x + c.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Indefinite Integration - Miscellaneous Exercise 3 [पृष्ठ १५०]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Indefinite Integration
Miscellaneous Exercise 3 | Q 1.19 | पृष्ठ १५०

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

Integrate the functions:

`(sin^(-1) x)/(sqrt(1-x^2))`


Integrate the functions:

`(2cosx - 3sinx)/(6cos x + 4 sin x)`


Integrate the functions:

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


Integrate the functions:

`(x^3 sin(tan^(-1) x^4))/(1 + x^8)`


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


Evaluate: `int (2y^2)/(y^2 + 4)dx`


\[\int\sqrt{x - x^2} dx\]

\[\int\sqrt{1 + x - 2 x^2} \text{ dx }\]

Write a value of

\[\int e^{\text{ log  sin x  }}\text{ cos x}. \text{ dx}\]

Write a value of

\[\int\frac{1 + \log x}{3 + x \log x} \text{ dx }\] .

Write a value of\[\int e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) dx\] .


Integrate the following functions w.r.t. x : `((sin^-1 x)^(3/2))/(sqrt(1 - x^2)`


Integrate the following functions w.r.t. x : `e^x.log (sin e^x)/tan(e^x)`


Integrate the following functions w.r.t. x : `(1)/(4x + 5x^-11)`


Integrate the following functions w.r.t. x : `sqrt(tanx)/(sinx.cosx)`


Integrate the following functions w.r.t. x : `(1)/(2 + 3tanx)`


Integrate the following functions w.r.t. x : `(20 + 12e^x)/(3e^x + 4)`


Integrate the following functions w.r.t. x : sin5x.cos8x


Evaluate the following:

`int sinx/(sin 3x)  dx`


Integrate the following functions w.r.t. x : `int (1)/(4 - 5cosx).dx`


Integrate the following functions w.r.t. x : `int (1)/(3 - 2cos 2x).dx`


Evaluate the following integrals:

`int (2x + 1)/(x^2 + 4x - 5).dx`


Evaluate the following integrals:

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


Choose the correct options from the given alternatives :

`int sqrt(cotx)/(sinx*cosx)*dx` =


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


If f '(x) = `"x"^2/2 - "kx" + 1`, f(0) = 2 and f(3) = 5, find f(x).


Evaluate: `int (2"e"^"x" - 3)/(4"e"^"x" + 1)` dx


`int cos sqrtx` dx = _____________


`int (log x)/(log ex)^2` dx = _________


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


`int sqrt(x)  sec(x)^(3/2) tan(x)^(3/2)"d"x`


`int1/(4 + 3cos^2x)dx` = ______ 


If `int x^3"e"^(x^2) "d"x = "e"^(x^2)/2 "f"(x) + "c"`, then f(x) = ______.


The integral `int ((1 - 1/sqrt(3))(cosx - sinx))/((1 + 2/sqrt(3) sin2x))dx` is equal to ______.


Write `int cotx  dx`.


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


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


Evaluate:

`int(cos 2x)/sinx dx`


`int (cos4x)/(sin2x + cos2x)dx` = ______.


Evaluate the following:

`int (1) / (x^2 + 4x - 5) dx`


Evaluate the following:

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


Evaluate:

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


Evaluate the following.

`int 1/ (x^2 + 4x - 5) dx`


Evaluate the following.

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


Evaluate:

`intsqrt(sec  x/2 - 1)dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×