हिंदी

Show that ∫0π2sin2xsinx+cosx=12log(2+1)

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

We have I = `int_0^(pi/2) (sin^2x)/(sinx + cosx)  "d"x`

= `int_0^(pi/2) (sin^2(pi/2 - x))/(sin(pi/2 - x) + cos(pi/2 - x)) "d"x`  ....(By P4)

⇒ I = `int_0^(pi/2) (cos^2x)/(sinx + cosx) "d"x`

Thus, we get 2I = `1/sqrt(2)  int_0^(pi/2)  ("d"x)/(cos(x - pi/4))`

= `1/sqrt(2) int_0^(pi/2) sec(x - pi/2) "d"x`

= `1/sqrt(2) [log(sec(x - pi/4) + tan(x - pi/4))]_0^(pi/2)`

= `1/sqrt(2)[log(sec  pi/4 + tan  pi/4) - log sec(- pi/4) + tan(- pi/4)]`

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

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

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

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

Hence I = `1/sqrt(2) log(sqrt(2) + 1)`.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Integrals - Solved Examples [पृष्ठ १५५]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 7 Integrals
Solved Examples | Q 17 | पृष्ठ १५५

संबंधित प्रश्न

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


 
 

Evaluate : `intlogx/(1+logx)^2dx`

 
 

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

`int_0^(pi/2) cos^2 x dx`


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

`int_0^(pi/2) sin^(3/2)x/(sin^(3/2)x + cos^(3/2) x) dx`


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

`int_(-5)^5 | x + 2| 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_0^pi (x  dx)/(1+ sin x)`


`∫_4^9 1/sqrtxdx=`_____

(A) 1

(B) –2

(C) 2

(D) –1


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


`int_0^1 "e"^(2x) "d"x` = ______


`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`


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


`int_0^{pi/2}((3sqrtsecx)/(3sqrtsecx + 3sqrt(cosecx)))dx` = ______ 


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


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


`int_0^1 x tan^-1x  dx` = ______ 


`int_-2^1 dx/(x^2 + 4x + 13)` = ______


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


`int_0^9 1/(1 + sqrtx)` dx = ______ 


`int_0^(pi/2)  cos x "e"^(sinx)  "d"x` is equal to ______.


If `int_0^"a" 1/(1 + 4x^2) "d"x = pi/8`, then a = ______.


If `int_a^b x^3 dx` = 0, then `(x^4/square)_a^b` = 0

⇒ `1/4 (square - square)` = 0

⇒ b4 – `square` = 0

⇒ (b2 – a2)(`square` + `square`) = 0

⇒ b2 – `square` = 0 as a2 + b2 ≠ 0

⇒ b = ± `square`


Let f be a real valued continuous function on [0, 1] and f(x) = `x + int_0^1 (x - t)f(t)dt`. Then, which of the following points (x, y) lies on the curve y = f(x)?


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 ______.


The value of `int_((-1)/sqrt(2))^(1/sqrt(2)) (((x + 1)/(x - 1))^2 + ((x - 1)/(x + 1))^2 - 2)^(1/2)`dx is ______.


Let `int_0^∞ (t^4dt)/(1 + t^2)^6 = (3π)/(64k)` then k is equal to ______.


Let f be continuous periodic function with period 3, such that `int_0^3f(x)dx` = 1. Then the value of `int_-4^8f(2x)dx` is ______.


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


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


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


Solve the following.

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


Evaluate the following integral:

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


Evaluate the following integral:

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


Solve the following.

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


Evaluate:

`int_0^sqrt(2)[x^2]dx`


Evaluate the following integral:

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


Evaluate the following definite integral:

`int_-2^3(1)/(x + 5)  dx`


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×