मराठी

If ad∫0a11+4x2dx=π8, then a = ______. - Mathematics

Advertisements
Advertisements

प्रश्न

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

रिकाम्या जागा भरा

उत्तर

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

Explanation:

Given that `int_0^"a" 1/(1 + 4x^2) "d"x = pi/8`

⇒ `1/4 int_0^"a" 1/((1/4 + x^2)) "d"x = pi/8`

⇒ `int_0^pi 1/([(1/2)^2 + x^2]) "d"x = pi/2`

⇒ `1/(1/2) [tan^-1  x/(1/2)]_0^"a" = pi/2`

⇒ `2[tan^-1 2"a" - tan^-1 0] = pi/2`

⇒ `tan^-1 2"a" = pi/4`

⇒ 2a = `tan  pi/4`

⇒ 2a = 1

⇒ a = `1/2`.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Integrals - Exercise [पृष्ठ १६९]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
पाठ 7 Integrals
Exercise | Q 61 | पृष्ठ १६९

संबंधित प्रश्‍न

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_((-pi)/2)^(pi/2) sin^2 x  dx`


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


Evaluate : `int  "e"^(3"x")/("e"^(3"x") + 1)` dx


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


Evaluate the following integral:

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


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


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


f(x) =  `{:{(x^3/k;       0 ≤ x ≤ 2), (0;     "otherwise"):}` is a p.d.f. of X. The value of k is ______


`int_(pi/4)^(pi/2) sqrt(1-sin 2x)  dx =` ______.


`int_0^(pi/2) 1/(1 + cosx) "d"x` = ______.


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


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


Evaluate the following:

`int_0^(pi/2)  "dx"/(("a"^2 cos^2x + "b"^2 sin^2 x)^2` (Hint: Divide Numerator and Denominator by cos4x)


Evaluate: `int_0^(2π) (1)/(1 + e^(sin x)`dx


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


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`


`int_0^5 cos(π(x - [x/2]))dx` where [t] denotes greatest integer less than or equal to t, is equal to ______.


With the usual notation `int_1^2 ([x^2] - [x]^2)dx` is equal to ______.


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


Evaluate the following limit :

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


Evaluate the following definite integral:

`int_1^3 log x  dx`


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 integral:

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×