Advertisements
Advertisements
Question
Integrate the following w.r.t.x : cot–1 (1 – x + x2)
Advertisements
Solution
Let I = `int cot^-1 (1 - x + x^2)*dx`
= `int tan^-1 (1/(1 - x + x^2))*dx`
= `int tan^-1 [(x + (1 - x))/(1 - x(1 - x))]`
= `int [tan^-1 x + tan^-1 (1 - x)]*dx`
= `int tan^-1 x*dx + int tan^-1 (1 - x)*dx`
∴ I = I1 + I2 ...(1)
I1 = `int tan^-1 x*dx = int(tan^-1x)1*dx`
= `(tan^-1x)* int 1dx - [d/dx (tan^-1x)* int 1dx]*dx`
= `(tan^-1x)x - int 1/(1 + x^2)*x*dx`
= `xtan^-1 x - (1)/(2) int (2x)/(1 + x^2)*dx`
∴ I1 = `x tan^-1x - (1)/(2)log|1 + x^2| + c_1`
...`[because d/dx (1 + x^2) = 2x and int (f'(x))/f(x) dx = log|f(x)| + c]`
I2 = `int tan^-1 (1 - x)*dx`
= `int tan^-1 (1 - x)]*1dx`
= `[tan^-1 (1 - x)]*int 1dx - int {d/dx [tan^-1 (1 - x)]* int 1dx}*dx`
= `[tan^-1 (1 - x)]*x - int (1)/(1 + (1 - x)^2)*(-1)*xdx`
= `xtan^-1 (1 - x) + int x/(1 + 1 - 2x + x^2)*dx`
= `xtan^-1 (1 - x) + int x/(2 - 2x + x^2)*dx`
Let x = `"A"[d/dx (2 - 2x + x^2)] + "B"`
∴ x = A(– 2 + 2x) + B = 2Ax + (–2A + B)
Comparing the coefficient of x and constant on both the sides, we get
1 = 2A and 0 = – 2A + B
∴ A = `(1)/(2) and 0 = -2(1/2) + "B"`
∴ B = 1
∴ x = `(1)/(2)(- 2 + 2x) + 1`
∴ I2= `xtan^-1 (1 - x) + int (1/2(-2 + 2x) + 1)/(2 - 2x + x^2)*dx`
= `xtan^-1 (1 - x) + 1/2 (-2 + 2x)/(2 - 2x + x^2)*dx + int (1)/(2 - 2x + x^2)*dx`
= `xtan^-1 (1 - x) + (1)/(2) log|2 - 2x + x^2| + int (1)/(1 + (1 - 2x + x^2))*dx`
= `xtan^-1 (1 - x) + (1)/(2) log|x^2 - 2x + 2| + int (1)/(1 + (1 - x^2))*dx`
= `xtan^-1 (1 - x) + (1)/(2) log|x^2 - 2x + 2| + (1)/(1) (tan-1 (1 - x))/(-1) + c_2`
= `x tan^-1 (1 - x) + 1/2log|x^2 - 2x + 2| - tan^-1 (1 - x) + c_2`
= `(x - 1)tan^-1 (1 - x) + (1)/(2)log|x^2 - 2x + 2| + c_2`
∴ I2 = `-(1 - x)tan^-1 (1 - x) + (1)/(2)log|x^2 - 2x + 2| + c_2` ...(3)
From (1),(2) and (3), we get
I = `x tan^-1 x - (1)/(2) log|1 + x^2| + c_1 - (1 - x)tan^-1 (1 - x) + 1/2log|x^2 - 2x + 2| + c_2`
= `x tan^-1 x - (1)/(2) log|1 + x^2| - (1 - x)tan^-1 (1 - x) + 1/2 |x^2 - 2x + 2| + c`, where c = c1 + c2.
APPEARS IN
RELATED QUESTIONS
Integrate : sec3 x w. r. t. x.
`int1/xlogxdx=...............`
(A)log(log x)+ c
(B) 1/2 (logx )2+c
(C) 2log x + c
(D) log x + c
Integrate the function in x sin x.
Integrate the function in x sin 3x.
Integrate the function in (sin-1x)2.
Integrate the function in (x2 + 1) log x.
Integrate the function in `sin^(-1) ((2x)/(1+x^2))`.
Find :
`∫(log x)^2 dx`
Evaluate the following : `int x^3.logx.dx`
Evaluate the following : `int e^(2x).cos 3x.dx`
Evaluate the following: `int x.sin^-1 x.dx`
Evaluate the following:
`int x.sin 2x. cos 5x.dx`
Integrate the following functions w.r.t. x : `x^2 .sqrt(a^2 - x^6)`
Integrate the following functions w.r.t. x : `sec^2x.sqrt(tan^2x + tan x - 7)`
Choose the correct options from the given alternatives :
`int (log (3x))/(xlog (9x))*dx` =
Integrate the following with respect to the respective variable : `t^3/(t + 1)^2`
Integrate the following with respect to the respective variable : `(3 - 2sinx)/(cos^2x)`
Integrate the following w.r.t.x : `sqrt(x)sec(x^(3/2))*tan(x^(3/2))`
Evaluate the following.
`int x^2 e^4x`dx
Evaluate the following.
`int "x"^3 "e"^("x"^2)`dx
Evaluate the following.
`int e^x (1/x - 1/x^2)`dx
Evaluate the following.
`int "e"^"x" "x - 1"/("x + 1")^3` dx
Evaluate: `int e^x/sqrt(e^(2x) + 4e^x + 13)` dx
`int (sin(x - "a"))/(cos (x + "b")) "d"x`
`int sqrt(tanx) + sqrt(cotx) "d"x`
`int 1/(x^2 - "a"^2) "d"x` = ______ + c
`int logx/(1 + logx)^2 "d"x`
∫ log x · (log x + 2) dx = ?
Evaluate the following:
`int (sin^-1 x)/((1 - x)^(3/2)) "d"x`
`int x/((x + 2)(x + 3)) dx` = ______ + `int 3/(x + 3) dx`
If `int(x + (cos^-1 3x)^2)/sqrt(1 - 9x^2)dx = 1/α(sqrt(1 - 9x^2) + (cos^-1 3x)^β) + C`, where C is constant of integration , then (α + 3β) is equal to ______.
If `π/2` < x < π, then `intxsqrt((1 + cos2x)/2)dx` = ______.
`int e^x [(2 + sin 2x)/(1 + cos 2x)]dx` = ______.
`int((4e^x - 25)/(2e^x - 5))dx = Ax + B log(2e^x - 5) + c`, then ______.
If `int (f(x))/(log(sin x))dx` = log[log sin x] + c, then f(x) is equal to ______.
Find: `int e^(x^2) (x^5 + 2x^3)dx`.
`int1/sqrt(x^2 - a^2) dx` = ______
Evaluate the following.
`int x^3 e^(x^2) dx`
`inte^(xloga).e^x dx` is ______
`int logx dx = x(1+logx)+c`
If u and v are two differentiable functions of x, then prove that `intu*v*dx = u*intv dx - int(d/dx u)(intv dx)dx`. Hence evaluate: `intx cos x dx`
If f'(x) = 4x3 - 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x)
Evaluate:
`inte^x "cosec" x(1 - cot x)dx`
Evaluate the following.
`intx^3 e^(x^2)dx`
