Advertisements
Advertisements
प्रश्न
Evaluate: `int_0^(π/4) log(1 + tanx)dx`.
Advertisements
उत्तर
Let I = `int_0^(π/4) log_e (1 + tan x)dx` ...(i)
`\implies` I = `int_0^(π/4) log_e (1 + tan(π/4 - x))dx`,
Using `int_0^a f(x)dx = int_0^a f(a - x)dx`
`\implies` I = `int_0^(π/4) log_e (1 + (1 - tanx)/(1 + tanx))dx`
= `int_0^(π/4) log_e (2/(1 + tanx))dx`
= `int_0^(π/4) log_e 2dx - I` ...(Using ...(i))
`\implies` 2I = `π/4 log_e 2`
`\implies` I = `π/8 log_e 2`.
APPEARS IN
संबंधित प्रश्न
Evaluate : `int e^x[(sqrt(1-x^2)sin^-1x+1)/(sqrt(1-x^2))]dx`
If `int_0^alpha(3x^2+2x+1)dx=14` then `alpha=`
(A) 1
(B) 2
(C) –1
(D) –2
By using the properties of the definite integral, evaluate the integral:
`int_(-5)^5 | x + 2| dx`
The total revenue R = 720 - 3x2 where x is number of items sold. Find x for which total revenue R is increasing.
Evaluate the following integral:
`int_0^1 x(1 - x)^5 *dx`
`int_0^2 e^x dx` = ______.
Evaluate `int_1^2 (sqrt(x))/(sqrt(3 - x) + sqrt(x)) "d"x`
`int_-9^9 x^3/(4 - x^2)` dx = ______
`int_0^{pi/2} xsinx dx` = ______
`int_0^(pi/2) sqrt(cos theta) * sin^2 theta "d" theta` = ______.
The value of `int_2^7 (sqrtx)/(sqrt(9 - x) + sqrtx)dx` is ______
`int_0^pi x sin^2x dx` = ______
`int_0^1 "e"^(5logx) "d"x` = ______.
Find `int_0^(pi/4) sqrt(1 + sin 2x) "d"x`
`int_0^(2"a") "f"(x) "d"x = 2int_0^"a" "f"(x) "d"x`, if f(2a – x) = ______.
Evaluate: `int_0^(2π) (1)/(1 + e^(sin x)`dx
The integral `int_0^2||x - 1| -x|dx` is equal to ______.
The value of the integral `int_0^sqrt(2)([sqrt(2 - x^2)] + 2x)dx` (where [.] denotes greatest integer function) is ______.
Evaluate `int_-1^1 |x^4 - x|dx`.
Evaluate: `int_0^π x/(1 + sinx)dx`.
For any integer n, the value of `int_-π^π e^(cos^2x) sin^3 (2n + 1)x dx` is ______.
If `int_0^1(3x^2 + 2x+a)dx = 0,` then a = ______
Evaluate `int_1^2(x+3)/(x(x+2)) dx`
`int_-9^9 x^3/(4-x^2) dx` =______
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`
Evaluate:
`int_0^6 |x + 3|dx`
\[\int_{-2}^{2}\left|x^{2}-x-2\right|\mathrm{d}x=\]
`∫_0^(π/2) (sqrttan x + sqrtcot x)dx` = ______.
