हिंदी

Evaluate: ∫π6π3dx1+tanx

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

Let I = `int_(pi/6)^(pi/3) (dx)/(1 + sqrt(tanx)`

= `int_(pi/6)^(pi/3) sqrt(cosx)/(sqrt(sinx) + sqrt(cos x)) dx`  ......(i)

Using `int_a^b f(x) dx = int_a^b f(a + b - x) dx`

I = `int_(pi/6)^(pi/3) sqrt(cos(pi/6 + pi/3 - x))/(sqrt(sin(pi/6 + pi/3 - x)) + sqrt(cos(pi/6 + pi/3 - x)))`

I = `int_(pi/6)^(pi/3) sqrt(sinx)/(sqrt(cosx) + sqrt(sinx)) dx`  ......(ii)

Adding (i) and (ii), we get

2I = `int_(pi/6)^(pi/3) sqrt(cosx)/(sqrt(sinx) + sqrt(cosx)) dx + int_(pi/6)^(pi/3) sqrt(sinx)/(sqrt(cosx) + sqrt(sinx)) dx`

2I = `int_(pi/6)^(pi/3) dx`

= `[x]_(pi/6)^(pi/3)`

= `pi/3 - pi/6`

= `pi/6`

Hence, I = `int_(pi/6)^(pi/3) (dx)/(1 + sqrt(tanx)) = pi/12`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2022-2023 (March) Sample

संबंधित प्रश्न

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/4) log (1+ tan x) dx`


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

`int_0^pi (x  dx)/(1+ sin x)`


Prove that `int _a^b f(x) dx = int_a^b f (a + b -x ) dx`  and hence evaluate   `int_(pi/6)^(pi/3) (dx)/(1 + sqrt(tan x))` .   


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


Evaluate: `int_0^pi ("x"sin "x")/(1+ 3cos^2 "x") d"x"`.


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


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)` log(1 + tanθ) dθ = ______


`int_3^9 x^3/((12 - x)^3 + x^3)` dx = ______ 


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


`int_{pi/6}^{pi/3} sin^2x dx` = ______ 


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


`int_0^pi x sin^2x dx` = ______ 


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


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


Evaluate the following:

`int_(-pi/4)^(pi/4) log|sinx + cosx|"d"x`


`int (dx)/(e^x + e^(-x))` is equal to ______.


If `f(a + b - x) = f(x)`, then `int_0^b x f(x)  dx` is equal to


Evaluate: `int_0^(π/2) 1/(1 + (tanx)^(2/3)) dx`


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 integral `int_0^2||x - 1| -x|dx` is equal to ______.


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


Evaluate the following limit :

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


Evaluate the following integrals:

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


Solve.

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


Evaluate the following integral:

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


Evaluate:

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×