हिंदी

Evaluate π∫0π/4log(1+tanx)dx.

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

Let I = `int_0^(π//4) log (1 + tanx)dx`  ...(i)

By using property

`int_0^a f(x) = int_0^a f(a - x)`

I = `int_0^(π//4) log [1 + tan(π/4 - x)]dx`

= `int_0^(π//4) log [1 + (tan  π/4 - tan x)/(1 + tan  π/4 tan x)]dx`

= `int_0^(π//4) log [1 + (1 - tanx)/(1 + tanx)]dx`

= `int_0^(π//4) log [2/(1 + tanx)]dx`

= `int_0^(π//4) log 2 - int_0^(π//4) log (1 + tanx)dx`  ...(ii)

On adding equations (i) and (ii),

2I = `int_0^(π//4) log (1 + tanx)dx + int_0^(π//4) log 2 dx - int_0^(π//4) log (1 + tanx)dx`

`\implies` 2I = `int_0^(π//4) log 2 dx` 

`\implies` 2I = `log 2 int_0^(π//4) 1.dx`

`\implies` 2I = `log 2 [x]_0^(π//4)`

`\implies` I = `log2/2 xx π/4 = π/8 log 2`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2022-2023 (March) Outside Delhi Set 1

संबंधित प्रश्न

If `int_0^alpha3x^2dx=8` then the value of α is :

(a) 0

(b) -2

(c) 2 

(d) ±2


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

`int_((-pi)/2)^(pi/2) sin^2 x  dx`


Show that `int_0^a f(x)g (x)dx = 2 int_0^a f(x) dx`  if f and g are defined as f(x) = f(a-x) and g(x) + g(a-x) = 4.


`int_(-pi/2)^(pi/2) (x^3 + x cos x + tan^5 x + 1) dx ` is ______.


\[\int\limits_0^k \frac{1}{2 + 8 x^2} dx = \frac{\pi}{16},\] find the value of k.


Find `dy/dx, if y = cos^-1 ( sin 5x)`


Choose the correct alternative:

`int_(-9)^9 x^3/(4 - x^2)  "d"x` =


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


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


`int_0^1 ((x^2 - 2)/(x^2 + 1))`dx = ?


The c.d.f, F(x) associated with p.d.f. f(x) = 3(1- 2x2). If 0 < x < 1 is k`(x - (2x^3)/"k")`, then value of k is ______.


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


`int_(pi/18)^((4pi)/9) (2 sqrt(sin x))/(sqrt (sin x) + sqrt(cos x))` dx = ?


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


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


`int_0^1 "e"^(5logx) "d"x` = ______.


Find `int_0^(pi/4) sqrt(1 + sin 2x) "d"x`


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


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


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


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


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


Evaluate: `int_1^3 sqrt(x + 5)/(sqrt(x + 5) + sqrt(9 - x))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 x/(1 + sinx)dx`.


Solve the following.

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


The area enclosed between the graph of y = x3 and the lines x = 0, y = 1, y = 8 is ______.


`int_0^(pi/4) (cos^2 x)/(cos^2 x + 4 sin^2 x) dx` =


If \[f(-x)=-f(x)\], what is \[\int_{-a}^{a} f(x)\,dx\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×