Advertisements
Advertisements
प्रश्न
Evaluate the following : `int x^3.tan^-1x.dx`
उत्तर
Let I = `int x^3.tan^-1x.dx`
= `int (tan^-1 x).x^3dx`
= `(tan^-1x) int x^3.dx - int [{d/dx (tan^-1 x) int x^3.dx}].dx`
= `(tan^-1x) (x^4/4) - int (1/(1 + x^2))x^4/(4).dx`
= `x^4/(4) tan^-1x - (1)/(4) ((x^4 - 1) + 1)/(x^2 + 1)`
= `x^4/(4) tan^-1x - (1)/(4) int ((x^2 - 1)(x^2 + 1) + 1)/(x^2 + 1).dx`
= `x^4/(4) tan^-1x - (1)/(4) int [x^2 - 1 + 1/(x^2 + 1)].dx`
= `x^4/(4) tan^-1x - (1)/(4) int [int x^2.dx - int 1.dx + int 1/(x^2 + 1).dx]`
= `x^4/(4) tan^-1x - (1)/(4)[x^3/3 - x + tan^-1x] + c`
= `x^4/(4) tan^-1x - tan^-1 x/(4) - x^3/(12) - x/(4) + c`
= `(1)/(4) (tan^-1x) (x^4 - 1) - x/(12) (x^2 - 3) + c`.
APPEARS IN
संबंधित प्रश्न
If `int_(-pi/2)^(pi/2)sin^4x/(sin^4x+cos^4x)dx`, then the value of I is:
(A) 0
(B) π
(C) π/2
(D) π/4
Integrate the function in x sin x.
Integrate the function in `x^2e^x`.
Integrate the function in x log x.
Integrate the function in x log 2x.
Integrate the function in x2 log x.
Integrate the function in x tan-1 x.
Integrate the function in `(x cos^(-1) x)/sqrt(1-x^2)`.
Integrate the function in `e^x (1/x - 1/x^2)`.
Integrate the function in e2x sin x.
Find :
`∫(log x)^2 dx`
Evaluate the following:
`int sec^3x.dx`
Evaluate the following : `int (t.sin^-1 t)/sqrt(1 - t^2).dt`
Evaluate the following : `int cos sqrt(x).dx`
Evaluate the following:
`int x.sin 2x. cos 5x.dx`
Integrate the following functions w.r.t. x : `e^(2x).sin3x`
Integrate the following functions w.r.t. x : `sec^2x.sqrt(tan^2x + tan x - 7)`
Integrate the following functions w.r.t. x : `((1 + sin x)/(1 + cos x)).e^x`
Integrate the following functions w.r.t. x : `e^x .(1/x - 1/x^2)`
Integrate the following functions w.r.t. x : `e^x/x [x (logx)^2 + 2 (logx)]`
Integrate the following functions w.r.t. x : `e^(sin^-1x)*[(x + sqrt(1 - x^2))/sqrt(1 - x^2)]`
Integrate the following functions w.r.t. x : `log(1 + x)^((1 + x)`
Choose the correct options from the given alternatives :
`int (log (3x))/(xlog (9x))*dx` =
Choose the correct options from the given alternatives :
`int (1)/(cosx - cos^2x)*dx` =
Integrate the following with respect to the respective variable : `t^3/(t + 1)^2`
Integrate the following with respect to the respective variable : `(sin^6θ + cos^6θ)/(sin^2θ*cos^2θ)`
Integrate the following w.r.t.x : `sqrt(x)sec(x^(3/2))*tan(x^(3/2))`
Integrate the following w.r.t.x : `log (1 + cosx) - xtan(x/2)`
Integrate the following w.r.t.x : e2x sin x cos x
Integrate the following w.r.t.x : sec4x cosec2x
Choose the correct alternative from the following.
`int (("x"^3 + 3"x"^2 + 3"x" + 1))/("x + 1")^5 "dx"` =
Evaluate: `int "dx"/sqrt(4"x"^2 - 5)`
Evaluate: `int e^x/sqrt(e^(2x) + 4e^x + 13)` dx
Evaluate: `int "dx"/("x"[(log "x")^2 + 4 log "x" - 1])`
Evaluate: `int "dx"/(25"x" - "x"(log "x")^2)`
Evaluate: ∫ (log x)2 dx
`int (sinx)/(1 + sin x) "d"x`
`int 1/sqrt(2x^2 - 5) "d"x`
`int ["cosec"(logx)][1 - cot(logx)] "d"x`
`int ("d"x)/(x - x^2)` = ______
State whether the following statement is True or False:
If `int((x - 1)"d"x)/((x + 1)(x - 2))` = A log|x + 1| + B log|x – 2|, then A + B = 1
Evaluate `int 1/(x log x) "d"x`
Evaluate `int 1/(4x^2 - 1) "d"x`
`int "e"^x x/(x + 1)^2 "d"x`
`int_0^"a" sqrt("x"/("a" - "x")) "dx"` = ____________.
Evaluate the following:
`int (sin^-1 x)/((1 - x)^(3/2)) "d"x`
State whether the following statement is true or false.
If `int (4e^x - 25)/(2e^x - 5)` dx = Ax – 3 log |2ex – 5| + c, where c is the constant of integration, then A = 5.
Find: `int (2x)/((x^2 + 1)(x^2 + 2)) dx`
If `π/2` < x < π, then `intxsqrt((1 + cos2x)/2)dx` = ______.
`int((4e^x - 25)/(2e^x - 5))dx = Ax + B log(2e^x - 5) + c`, then ______.
Find `int (sin^-1x)/(1 - x^2)^(3//2) dx`.
Find: `int e^(x^2) (x^5 + 2x^3)dx`.
Evaluate :
`int(4x - 6)/(x^2 - 3x + 5)^(3/2) dx`
`intsqrt(1+x) dx` = ______
Solution of the equation `xdy/dx=y log y` is ______
Evaluate the following.
`int x^3 e^(x^2) dx`
`int(3x^2)/sqrt(1+x^3) dx = sqrt(1+x^3)+c`
Solve the differential equation (x2 + y2) dx - 2xy dy = 0 by completing the following activity.
Solution: (x2 + y2) dx - 2xy dy = 0
∴ `dy/dx=(x^2+y^2)/(2xy)` ...(1)
Puty = vx
∴ `dy/dx=square`
∴ equation (1) becomes
`x(dv)/dx = square`
∴ `square dv = dx/x`
On integrating, we get
`int(2v)/(1-v^2) dv =intdx/x`
∴ `-log|1-v^2|=log|x|+c_1`
∴ `log|x| + log|1-v^2|=logc ...["where" - c_1 = log c]`
∴ x(1 - v2) = c
By putting the value of v, the general solution of the D.E. is `square`= cx
`int logx dx = x(1+logx)+c`
`int(f'(x))/sqrt(f(x)) dx = 2sqrt(f(x))+c`
Evaluate:
`int e^(logcosx)dx`
Evaluate the following.
`intx^3 e^(x^2) dx`
Evaluate:
`int1/(x^2 + 25)dx`
Evaluate `int (1 + x + x^2/(2!))dx`
If f'(x) = 4x3 - 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x)
Evaluate:
`int x^2 cos x dx`
Evaluate the following.
`int x sqrt(1 + x^2) dx`
Evaluate:
`inte^x "cosec" x(1 - cot x)dx`
Evaluate `int(1 + x + x^2/(2!))dx`.