मराठी

D∫0π21-sin2x dx is equal to ______.

Advertisements
Advertisements

प्रश्न

`int_0^(pi/2) sqrt(1 - sin2x)  "d"x` is equal to ______.

पर्याय

  • `2sqrt(2)`

  • `2(sqrt(2) + 1)`

  • 2

  • `2(sqrt(2) - 1)`

MCQ
रिकाम्या जागा भरा
Advertisements

उत्तर

`int_0^(pi/2) sqrt(1 - sin2x)  "d"x` is equal to `2(sqrt(2) - 1)`.

Explanation:

Let I = `int_0^(pi/2) sqrt(1 - sin2x)  "d"x`

= `int_0^(pi/2) sqrt((sin^2x + cos^2x - 2 sinx cosx))  "d"x`

= `int_0^(pi/2) sqrt((sinx - cosx)^2)  "d"x`

= `int_0^(pi/2) +- (sinx - cosx)  "d"x`

= `int_0^(pi/4) - (sin x - cosx)  "d"x + int_(pi/4)^(pi/2) (sinx - cosx)  "dx`

= `int_0^(pi/4) (cosx - sinx)  "d"x + int_(pi/4)^(pi/2) (sinx - cosx)  "d"x`

= `[sinx + cosx]_0^(pi/4) + [- cosx - sinx]_(pi/4)^(pi/2)`

= `[(sin  pi/4 + cos  pi/4) - (sin0 - cos0)] - [(cos  pi/2 + sin  pi/2) - (cos  pi/4 + sin  pi/4)]`

= `[(1/sqrt(2) + 1/sqrt(2)) - (+ 1)] - [(0 + 1) - (1/sqrt(2) + 1/sqrt(2))]`

= `(2/sqrt(2) - 1) - (1 - 2/sqrt(2))`

= `2/sqrt(2) - 1 -1 + 2/(sqrt(2))`

= `4/sqrt(2) - 2`

= `2sqrt(2) - 2`

= `2(sqrt(2) - 1)`.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Integrals - Exercise [पृष्ठ १६९]

APPEARS IN

संबंधित प्रश्‍न

If `int_0^alpha3x^2dx=8` then the value of α is :

(a) 0

(b) -2

(c) 2 

(d) ±2


By using the properties of the definite integral, evaluate the integral:

`int_0^(pi/2)  sqrt(sinx)/(sqrt(sinx) + sqrt(cos x)) dx` 


By using the properties of the definite integral, evaluate the integral:

`int_0^1 x(1-x)^n dx`


By using the properties of the definite integral, evaluate the integral:

`int_(pi/2)^(pi/2) sin^7 x dx`


By using the properties of the definite integral, evaluate the integral:

`int_0^(pi/2) (sin x - cos x)/(1+sinx cos x) dx`


Show that `int_0^a f(x)g (x)dx = 2 int_0^a f(x) dx`  if f and g are defined as f(x) = f(a-x) and g(x) + g(a-x) = 4.


The value of `int_0^(pi/2) log  ((4+ 3sinx)/(4+3cosx))` dx is ______.


Evaluate: `int_1^4 {|x -1|+|x - 2|+|x - 4|}dx`


Evaluate : \[\int(3x - 2) \sqrt{x^2 + x + 1}dx\] .


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


Evaluate = `int (tan x)/(sec x + tan x)` . dx


Evaluate the following integrals : `int_2^5 sqrt(x)/(sqrt(x) + sqrt(7 - x))*dx`


State whether the following statement is True or False:

`int_(-5)^5 x/(x^2 + 7)  "d"x` = 10


`int (cos x + x sin x)/(x(x + cos x))`dx = ?


`int_-9^9 x^3/(4 - x^2)` dx = ______


`int_0^pi x*sin x*cos^4x  "d"x` = ______.


`int_0^pi x sin^2x dx` = ______ 


`int_0^9 1/(1 + sqrtx)` dx = ______ 


Evaluate `int_0^(pi/2) (tan^7x)/(cot^7x + tan^7x) "d"x`


`int_a^b f(x)dx` = ______.


If `intxf(x)dx = (f(x))/2` then f(x) = ex.


Let a be a positive real number such that `int_0^ae^(x-[x])dx` = 10e – 9 where [x] is the greatest integer less than or equal to x. Then, a is equal to ______.


If f(x) = `(2 - xcosx)/(2 + xcosx)` and g(x) = logex, (x > 0) then the value of the integral `int_((-π)/4)^(π/4) "g"("f"(x))"d"x` is ______.


`int_0^1|3x - 1|dx` equals ______.


If `int_0^K dx/(2 + 18x^2) = π/24`, then the value of K is ______.


`int_-1^1 (17x^5 - x^4 + 29x^3 - 31x + 1)/(x^2 + 1) dx` is equal to ______.


If `int_0^(2π) cos^2 x  dx = k int_0^(π/2) cos^2 x  dx`, then the value of k is ______.


Evaluate the following limit :

`lim_("x"->3)[sqrt("x"+6)/"x"]`


Evaluate the following definite integral:

`int_4^9 1/sqrt"x" "dx"`


Evaluate the following integral:

`int_0^1 x(1-x)^5 dx`


Evaluate `int_1^2(x+3)/(x(x+2))  dx`


Evaluate the following definite integral:

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


Evaluate the following integral:

`int_-9^9 x^3/(4 - x^2) dx`


Evaluate the following integral:

`int_-9^9x^3/(4-x^2)dx`


Evaluate the following integral:

`int_0^1 x (1 - x)^5 dx`


Evaluate the following integral:

`int_0^1x(1-x)^5dx`


`∫_0^(π/2) (sqrttan x + sqrtcot x)dx` = ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×