Advertisements
Advertisements
प्रश्न
Show that : `int _0^(pi/4) "log" (1+"tan""x")"dx" = pi /8 "log"2`
Advertisements
उत्तर
Let I = `int _0^(pi/4) "log"(1+"tan""x")"dx"`
= `int _0^(pi/4) "log"(1+ "tan""x")"dx"`
`=int _0^(pi/4) "log"{1+"tan"(pi/4-"x")} "dx"`
`(because int _0^"a" "f" ("x") "dx" int "f"("a" -"x")"dx")`
`=int _0^(pi/4)"log"{1+(("tan"pi/4 - "tan""x"))/(1+"tan"pi/4"tan""x")} "dx"`
`=int _0^(pi/4) "log"{1+(1-"tan""x")/(1+ "tan""x")} "dx"`
`=int _0^(pi/4) "log"{(1 + "tan""x" +1 -"tan""x")/(1 + "tan""x")}"dx"`
`=int _0^(pi/4) "log"(2/(1+"tan""x")) "dx"`
`=int _0^(pi/4) {"log" 2 -"log"(1+ "tan""x")} "dx"`
`=int _0^(pi/4) "log"2"dx" - int _0^(pi/4) "log" (1+"tan""x")"dx"`
`"I" = "log"2["x"]int _0^(pi/4) - "I"`
2I = `"log" 2 [pi/4-0]`
`"I" = pi/8 ."log"2`
` therefore int _0^(pi/4) "log"(1 +"tan""x")"dx" = pi/8"log"2`
APPEARS IN
संबंधित प्रश्न
Find : `int((2x-5)e^(2x))/(2x-3)^3dx`
Evaluate :
`int1/(sin^4x+sin^2xcos^2x+cos^4x)dx`
Integrate the functions:
`(x^3 - 1)^(1/3) x^5`
Solve:
dy/dx = cos(x + y)
Evaluate: `int_0^3 f(x)dx` where f(x) = `{(cos 2x, 0<= x <= pi/2),(3, pi/2 <= x <= 3) :}`
Write a value of
Write a value of\[\int\frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} dx\]
Write a value of\[\int\sqrt{9 + x^2} \text{ dx }\].
Evaluate the following integrals:
`int (cos2x)/sin^2x dx`
Evaluate the following integrals : `int (cos2x)/(sin^2x.cos^2x)dx`
Integrate the following functions w.r.t. x : `(1 + x)/(x.sin (x + log x)`
Integrate the following functions w.r.t. x : `e^x.log (sin e^x)/tan(e^x)`
Integrate the following functions w.r.t. x : `sin(x - a)/cos(x + b)`
Integrate the following functions w.r.t. x : tan 3x tan 2x tan x
Evaluate the following : `int (1)/(4x^2 - 3).dx`
Evaluate the following : `int (1)/(cos2x + 3sin^2x).dx`
Integrate the following functions w.r.t. x : `int (1)/(3 + 2sinx).dx`
Evaluate the following integrals : `int sqrt((x - 7)/(x - 9)).dx`
Choose the correct options from the given alternatives :
`int (e^(2x) + e^-2x)/e^x*dx` =
Integrate the following w.r.t.x: `(3x + 1)/sqrt(-2x^2 + x + 3)`
Evaluate `int (-2)/(sqrt("5x" - 4) - sqrt("5x" - 2))`dx
Evaluate: If f '(x) = `sqrt"x"` and f(1) = 2, then find the value of f(x).
`int 1/sqrt((x - 3)(x + 2))` dx = ______.
`int sqrt(1 + sin2x) dx`
`int sqrt(("e"^(3x) - "e"^(2x))/("e"^x + 1)) "d"x`
Evaluate `int"e"^x (1/x - 1/x^2) "d"x`
`int x^3"e"^(x^2) "d"x`
If f'(x) = `x + 1/x`, then f(x) is ______.
If `int(cosx - sinx)/sqrt(8 - sin2x)dx = asin^-1((sinx + cosx)/b) + c`. where c is a constant of integration, then the ordered pair (a, b) is equal to ______.
`int (sin (5x)/2)/(sin x/2)dx` is equal to ______. (where C is a constant of integration).
If `int [log(log x) + 1/(logx)^2]dx` = x [f(x) – g(x)] + C, then ______.
Solve the following Evaluate.
`int(5x^2 - 6x + 3)/(2x - 3)dx`
Evaluate the following.
`int x^3/sqrt(1+x^4) dx`
Evaluate `int (1 + "x" + "x"^2/(2!))`dx
Evaluate the following.
`intx^3/sqrt(1+x^4)dx`
Which standard substitution is used for \[\sqrt{x^2-a^2}\], \[\frac{1}{\sqrt{x^2-a^2}}\], or \[x^2-a^2\]?
When applying substitution, what must always be rewritten in terms of the new variable?
