English

Evaluate: ∫dx2+cosx-sinx

Advertisements
Advertisements

Question

Evaluate: `int (dx)/(2 + cos x - sin x)`

Sum
Advertisements

Solution

Let I = `int (dx)/(2 + cos x - sin x)`

Put `tan  x/2` = t

⇒ x = 2 tan–1t

∴ dx = `(2 dt)/(1 + t^2)`

And sin x = `(2t)/(1 + t^2)`, cos x = `(1 - t^2)/(1 + t^2)`

∴ I = `int 1/(2 + ((1  -  t^2)/(1  +  t^2)) - ((2t)/(1 + t^2))) * (2dt)/(1  +  t^2)`

= `int (1 + t^2)/(2 + 2t^2 + 1 - t^2 - 2t) * (2dt)/(1 + t^2)`

= `2int 1/(t^2 - 2t + 3) dt`

= `2int 1/((t - 1)^2 + (sqrt(2))^2)  dt`

= `2 xx 1/sqrt(2)  tan^-1  ((t - 1)/sqrt(2)) + C`

= `sqrt(2)tan^-1 ((tan(x/2) - 1)/sqrt(2)) + C`

shaalaa.com
  Is there an error in this question or solution?
2021-2022 (March) Set 1

RELATED QUESTIONS

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


Evaluate:

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


Integrate the rational function:

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


Integrate the rational function:

`(3x + 5)/(x^3 - x^2 - x + 1)`


Integrate the rational function:

`(x^3 + x + 1)/(x^2 -1)`


Integrate the rational function:

`(3x -1)/(x + 2)^2`


Integrate the rational function:

`((x^2 +1)(x^2 + 2))/((x^2 + 3)(x^2+ 4))`


Integrate the rational function:

`1/(x(x^4 - 1))`


Evaluate : `∫(x+1)/((x+2)(x+3))dx`


Integrate the following w.r.t. x : `(2x)/(4 - 3x - x^2)`


Integrate the following w.r.t. x : `(x^2 + x - 1)/(x^2 + x - 6)`


Integrate the following w.r.t. x : `(1)/(sinx*(3 + 2cosx)`


Integrate the following w.r.t. x : `(2log x + 3)/(x(3 log x + 2)[(logx)^2 + 1]`


Choose the correct options from the given alternatives :

If `int tan^3x*sec^3x*dx = (1/m)sec^mx - (1/n)sec^n x + c, "then" (m, n)` =


Integrate the following w.r.t. x: `(2x^2 - 1)/(x^4 + 9x^2 + 20)`


Integrate the following with respect to the respective variable : `(cos 7x - cos8x)/(1 + 2 cos 5x)`


Integrate the following w.r.t.x : `x^2/sqrt(1 - x^6)`


Integrate the following w.r.t.x:

`x^2/((x - 1)(3x - 1)(3x - 2)`


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


Evaluate:

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


Evaluate:

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


For `int ("x - 1")/("x + 1")^3  "e"^"x" "dx" = "e"^"x"` f(x) + c, f(x) = (x + 1)2.


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


`int (2x - 7)/sqrt(4x- 1) dx`


`int "e"^(3logx) (x^4 + 1)^(-1) "d"x`


`int x^2sqrt("a"^2 - x^6)  "d"x`


If f'(x) = `x - 3/x^3`, f(1) = `11/2` find f(x)


`int (7 + 4x + 5x^2)/(2x + 3)^(3/2) dx`


`int 1/(4x^2 - 20x + 17)  "d"x`


`int sec^2x sqrt(tan^2x + tanx - 7)  "d"x`


`int (3x + 4)/sqrt(2x^2 + 2x + 1)  "d"x`


`int x sin2x cos5x  "d"x`


`int 1/(sinx(3 + 2cosx))  "d"x`


`int (3"e"^(2x) + 5)/(4"e"^(2x) - 5)  "d"x`


Choose the correct alternative:

`int (x + 2)/(2x^2 + 6x + 5) "d"x = "p"int (4x + 6)/(2x^2 + 6x + 5) "d"x + 1/2 int 1/(2x^2 + 6x + 5)"d"x`, then p = ?


If f'(x) = `1/x + x` and f(1) = `5/2`, then f(x) = log x + `x^2/2` + ______ + c


`int 1/(4x^2 - 20x + 17)  "d"x`


If `intsqrt((x - 7)/(x - 9)) dx = Asqrt(x^2 - 16x + 63) + log|x - 8 + sqrt(x^2 - 16x + 63)| + c`, then A = ______


Evaluate the following:

`int_"0"^pi  (x"d"x)/(1 + sin x)`


Evaluate the following:

`int "e"^(-3x) cos^3x  "d"x`


Evaluate the following:

`int sqrt(tanx)  "d"x`  (Hint: Put tanx = t2)


If `intsqrt((x - 5)/(x - 7))dx = Asqrt(x^2 - 12x + 35) + log|x| - 6 + sqrt(x^2 - 12x + 35) + C|`, then A = ______.


Evaluate:

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


If \[\int\frac{2x+3}{(x-1)(x^{2}+1)}\mathrm{d}x\] = \[=\log_{e}\left\{(x-1)^{\frac{5}{2}}\left(x^{2}+1\right)^{a}\right\}-\frac{1}{2}\tan^{-1}x+\mathrm{A}\] where A is an arbitrary constant, then the value of a is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×