हिंदी

Integrate the following w.r.t.x : log(1+cosx)-xtan(x2) - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

Let I = `int [log(1 + cosx) - xtan(x/2)]*dx`

= `int [log(1 + cos.x)*1dx - intxtan (x/2)*dx`

= `[log(1 + cosx)]* int 1dx - int {d/dx [log (1 + cosx)]* int 1dx}*dx - xtan (x/2)*dx`

= `[log (1 + cosx)]*(x) - int 1/(1 + cosx)*(0 - sin x)*xdx - int x tan (x/2)*dx`

= `x*log(1 + cosx) + intx* (sinx)/(1 + cosx)*dx - int xtan (x/2)*dx + c`

= `x*log(1 + cosx) + intx*(2sin(x/2)*cos(x/2))/(2cos^2(x/2)*dx - int xtan (x/2)*dx + c`

= `xlog (1 + cosx) + int x*tan(x/2)*dx - intxtan(x/2)*dx + c`

= x·log(1 + cosx) + 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 3.05 | पृष्ठ १५०

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Integrate the function in x tan-1 x.


Integrate the function in tan-1 x.


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


`intx^2 e^(x^3) dx` equals: 


Evaluate the following:

`int x^2 sin 3x  dx`


Evaluate the following : `int sin θ.log (cos θ).dθ`


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

`e^-x cos2x`


Integrate the following functions w.r.t. x : `sqrt(2x^2 + 3x + 4)`


Integrate the following functions w.r.t. x : [2 + cot x – cosec2x]e 


Integrate the following functions w.r.t. x : `[x/(x + 1)^2].e^x`


Choose the correct options from the given alternatives :

`int sin (log x)*dx` =


Integrate the following w.r.t.x : `(1)/(xsin^2(logx)`


Evaluate the following.

∫ x log x dx


Evaluate the following.

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


Evaluate the following.

`int e^x (1/x - 1/x^2)`dx


Evaluate: Find the primitive of `1/(1 + "e"^"x")`


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


Evaluate: `int "dx"/("9x"^2 - 25)`


Evaluate: `int "dx"/(5 - 16"x"^2)`


`int sin4x cos3x  "d"x`


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


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


`int"e"^(4x - 3) "d"x` = ______ + c


∫ log x · (log x + 2) dx = ?


`int cot "x".log [log (sin "x")] "dx"` = ____________.


Evaluate the following:

`int_0^1 x log(1 + 2x)  "d"x`


Evaluate the following:

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


The value of `int_(- pi/2)^(pi/2) (x^3 + x cos x + tan^5x + 1)  dx` is


The value of `int_0^(pi/2) log ((4 + 3 sin x)/(4 + 3 cos x))  dx` is


Evaluate: `int_0^(pi/4) (dx)/(1 + tanx)`


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


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


Evaluate :

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


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


Evaluate:

`intcos^-1(sqrt(x))dx`


`int (sin^-1 sqrt(x) + cos^-1 sqrt(x))dx` = ______.


Evaluate the following.

`intx^3  e^(x^2) dx`


Complete the following activity:

`int_0^2 dx/(4 + x - x^2) `

= `int_0^2 dx/(-x^2 + square + square)`

= `int_0^2 dx/(-x^2 + x + 1/4 - square + 4)`

= `int_0^2 dx/ ((x- 1/2)^2 - (square)^2)`

= `1/sqrt17 log((20 + 4sqrt17)/(20 - 4sqrt17))`


Evaluate the following:

`intx^3e^(x^2)dx` 


Evaluate the following. 

`int x sqrt(1 + x^2)  dx`  


Evaluate the following.

`int x^3 e^(x^2) dx` 


Evaluate the following.

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


Evaluate the following.

`intx^2e^(4x)dx`


Evaluate the following.

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×