हिंदी

D∫0π21-sin2x dx is equal to ______. - Mathematics

Advertisements
Advertisements

प्रश्न

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

विकल्प

  • `2sqrt(2)`

  • `2(sqrt(2) + 1)`

  • 2

  • `2(sqrt(2) - 1)`

MCQ
रिक्त स्थान भरें
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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Integrals - Exercise [पृष्ठ १६९]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 7 Integrals
Exercise | Q 58 | पृष्ठ १६९

संबंधित प्रश्न

Evaluate : `int e^x[(sqrt(1-x^2)sin^-1x+1)/(sqrt(1-x^2))]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^a  sqrtx/(sqrtx + sqrt(a-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.


Prove that `int_0^af(x)dx=int_0^af(a-x) dx`

hence evaluate `int_0^(pi/2)sinx/(sinx+cosx) dx`


\[\int_\pi^\frac{3\pi}{2} \sqrt{1 - \cos2x}dx\]

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


Evaluate  : `int "x"^2/("x"^4 + 5"x"^2 + 6) "dx"`


Find : `int_  (2"x"+1)/(("x"^2+1)("x"^2+4))d"x"`.


`int_1^2 1/(2x + 3)  dx` = ______


`int_2^4 x/(x^2 + 1)  "d"x` = ______


`int_0^(pi/4) (sec^2 x)/((1 + tan x)(2 + tan x))`dx = ?


`int_"a"^"b" sqrtx/(sqrtx + sqrt("a" + "b" - x)) "dx"` = ______.


If `int_0^"a" sqrt("a - x"/x) "dx" = "K"/2`, then K = ______.


`int_0^(pi/2) sqrt(cos theta) * sin^2 theta "d" theta` = ______.


`int_0^{pi/4} (sin2x)/(sin^4x + cos^4x)dx` = ____________


`int_0^{pi/2} (cos2x)/(cosx + sinx)dx` = ______


`int_0^1 "e"^(5logx) "d"x` = ______.


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


Show that `int_0^(pi/2) (sin^2x)/(sinx + cosx) = 1/sqrt(2) log (sqrt(2) + 1)`


`int_((-pi)/4)^(pi/4) "dx"/(1 + cos2x)` is equal to ______.


Evaluate:

`int_2^8 (sqrt(10 - "x"))/(sqrt"x" + sqrt(10 - "x")) "dx"`


Evaluate: `int_(pi/6)^(pi/3) (dx)/(1 + sqrt(tanx)`


`int_0^1 1/(2x + 5) dx` = ______.


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


Let `int ((x^6 - 4)dx)/((x^6 + 2)^(1/4).x^4) = (ℓ(x^6 + 2)^m)/x^n + C`, then `n/(ℓm)` is equal to ______.


`int_0^(π/4) x. sec^2 x  dx` = ______.


If `int_0^(π/2) log cos x  dx = π/2 log(1/2)`, then `int_0^(π/2) log sec dx` = ______.


`int_((-π)/2)^(π/2) log((2 - sinx)/(2 + sinx))` is equal to ______.


`int_-1^1 |x - 2|/(x - 2) dx`, x ≠ 2 is equal to ______.


For any integer n, the value of `int_-π^π e^(cos^2x) sin^3 (2n + 1)x  dx` is ______.


Evaluate:

`int_0^1 |2x + 1|dx`


Solve the following.

`int_0^1e^(x^2)x^3 dx`


Evaluate the following integral:

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


Evaluate:

`int_0^6 |x + 3|dx`


Evaluate the following definite integral:

`int_-2^3(1)/(x + 5)  dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×