मराठी

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

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` 

बेरीज
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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Integrals - Exercise 7.11 [पृष्ठ ३४७]

APPEARS IN

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

संबंधित प्रश्‍न

Prove that: `int_0^(2a)f(x)dx=int_0^af(x)dx+int_0^af(2a-x)dx`


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

`int_(-5)^5 | x + 2| dx`


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


Evaluate `int e^x [(cosx - sin x)/sin^2 x]dx`


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


If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that

\[\int_a^b xf\left( x \right)dx = \left( \frac{a + b}{2} \right) \int_a^b f\left( x \right)dx\]

The total revenue R = 720 - 3x2 where x is number of items sold. Find x for which total  revenue R is increasing.


`int_0^2 e^x dx` = ______.


`int_2^7 sqrt(x)/(sqrt(x) + sqrt(9 - x))  dx` = ______.


`int_(-7)^7 x^3/(x^2 + 7)  "d"x` = ______


State whether the following statement is True or False:

`int_(-5)^5 x/(x^2 + 7)  "d"x` = 10


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`


The value of `int_-3^3 ("a"x^5 + "b"x^3 + "c"x + "k")"dx"`, where a, b, c, k are constants, depends only on ______.


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


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


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


The value of `int_1^3 dx/(x(1 + x^2))` is ______ 


If `int_0^"k" "dx"/(2 + 32x^2) = pi/32,` then the value of k is ______.


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


`int_0^9 1/(1 + sqrtx)` dx = ______ 


Evaluate `int_(-1)^2 "f"(x)  "d"x`, where f(x) = |x + 1| + |x| + |x – 1|


`int_(-2)^2 |x cos pix| "d"x` is equal to ______.


`int_0^(pi/2) (sin^"n" x"d"x)/(sin^"n" x + cos^"n" x)` = ______.


`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_0^(π/2) 1/(1 + (tanx)^(2/3)) dx`


Evaluate: `int_1^3 sqrt(x)/(sqrt(x) + sqrt(4) - x) dx`


If f(x) = `{{:(x^2",", "where"  0 ≤ x < 1),(sqrt(x)",", "when"  1 ≤ x < 2):}`, then `int_0^2f(x)dx` equals ______.


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


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


Evaluate: `int_0^(π/4) log(1 + tanx)dx`.


Evaluate the following limit :

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


Evaluate the following integrals:

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


Evaluate the following integral:

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


Solve the following.

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×