हिंदी

By using the properties of the definite integral, evaluate the integral: ∫0π2 sinxsinx+cosxdx

Advertisements
Advertisements

प्रश्न

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

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

योग
Advertisements

उत्तर

Let `I = int_0^(pi/2) sqrtsinx/(sqrt sinx + sqrt cos x)  dx`     ...(i)

Replace x to `(pi/2 - x)` in (i)

`[∵ int_0^a f (x) dx = int_0^a f (a - x) dx]`

`I = int_0^(pi/2) (sqrt sin (pi/2 - x))/ (sqrt sin (pi/2 - x) + sqrt cos (pi/2 - x))  dx`

`I = int_0^(pi/2) sqrtcosx/(sqrtcos x + sqrt sin x)  dx`       ...(ii)

Adding (i) and (ii), we get

`2I = int_0^(pi/2) [sqrt sinx/ (sqrt sinx + sqrt cos x) + sqrt cos x/(sqrt cos x + sqrt sinx)]  dx` 

`= int_0^(pi/2) (sqrt cos x + sqrt sin x)/(sqrt cosx + sqrt sin x)`

`= int_0^(pi/2) dx = [x]_0^(pi/2)`

`= pi/2 - 0`

`= pi/2`

⇒ `I = pi/4`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Integrals - Exercise 7.11 [पृष्ठ ३४७]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 7 Integrals
Exercise 7.11 | Q 2 | पृष्ठ ३४७

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_0^(pi/2) (2log sin x - log sin 2x)dx`


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

`int_((-pi)/2)^(pi/2) sin^2 x  dx`


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

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


\[\int\limits_0^a 3 x^2 dx = 8,\] find the value of a.


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

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


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


`int_"a"^"b" "f"(x)  "d"x` = ______


`int_1^2 1/(2x + 3)  dx` = ______


Evaluate `int_1^3 x^2*log x  "d"x`


By completing the following activity, Evaluate `int_2^5 (sqrt(x))/(sqrt(x) + sqrt(7 - x))  "d"x`.

Solution: Let I = `int_2^5 (sqrt(x))/(sqrt(x) + sqrt(7 - x))  "d"x`     ......(i)

Using the property, `int_"a"^"b" "f"(x) "d"x = int_"a"^"b" "f"("a" + "b" - x)  "d"x`, we get

I = `int_2^5 ("(  )")/(sqrt(7 - x) + "(  )")  "d"x`   ......(ii)

Adding equations (i) and (ii), we get

2I = `int_2^5 (sqrt(x))/(sqrt(x) - sqrt(7 - x))  "d"x + (   )  "d"x`

2I = `int_2^5 (("(    )" + "(     )")/("(    )" + "(     )"))  "d"x`

2I = `square`

∴ I =  `square`


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


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


`int_0^(pi"/"4)` log(1 + tanθ) dθ = ______


`int_2^3 x/(x^2 - 1)` dx = ______


`int_0^{pi/2} xsinx 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_-2^1 dx/(x^2 + 4x + 13)` = ______


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


The value of `int_2^7 (sqrtx)/(sqrt(9 - x) + sqrtx)dx` is ______ 


Which of the following is true?


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


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


`int_0^(2"a") "f"("x") "dx" = int_0^"a" "f"("x") "dx" + int_0^"a" "f"("k" - "x") "dx"`, then the value of k is:


If `β + 2int_0^1x^2e^(-x^2)dx = int_0^1e^(-x^2)dx`, then the value of β is ______.


What is `int_0^(π/2)` sin 2x ℓ n (cot x) dx equal to ?


`int_(π/3)^(π/2) x sin(π[x] - x)dx` is equal to ______.


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


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"`


If `int_0^1(3x^2 + 2x+a)dx = 0,` then a = ______


Evaluate the following integral:

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


Evaluate the following integral:

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


Evaluate the following definite intergral:

`int_1^2 (3x)/(9x^2 - 1) dx`


Evaluate the following definite intergral:

`int_1^3logx  dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×