Advertisements
Advertisements
Question
If u and v ore differentiable functions of x. then prove that:
`int uv dx = u intv dx - int [(du)/(d) intv dx]dx`
Hence evaluate `intlog x dx`
Advertisements
Solution
Let `intv dx`= w ......(i)
Then, `(dw)/(dx)` = v ......(ii)
Now, `d/(dx)(u, w) = u. d/(dx)(w) + w d/(dx)(u)`
= `u.v + w (du)/(dx)` ......[From (ii)]
By definition of integration
u.w = `int[u.v + w . (du)/(dx)]dx`
∴ u.w = `intu.v dx + intw. (du)/(dx) dx`
∴ `intu.v dx = u.w - intw. (du)/(dx) dx`
= `uintv dx - int[(du)/(dx) intv. dx]dx` ......[Using (i)]
Hence, `intlogx dx = xlogx - int 1/x x xx dx`
= x log x – x + C
APPEARS IN
RELATED QUESTIONS
Integrate : sec3 x w. r. t. x.
Prove that:
`int sqrt(x^2 - a^2)dx = x/2sqrt(x^2 - a^2) - a^2/2log|x + sqrt(x^2 - a^2)| + c`
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
If u and v are two functions of x then prove that
`intuvdx=uintvdx-int[du/dxintvdx]dx`
Hence evaluate, `int xe^xdx`
Integrate the function in x cos-1 x.
Integrate the function in x sec2 x.
Integrate the function in tan-1 x.
Integrate the function in ex (sinx + cosx).
Integrate the function in `(xe^x)/(1+x)^2`.
Evaluate the following: `int x.sin^-1 x.dx`
Integrate the following functions w.r.t.x:
`e^-x cos2x`
Integrate the following functions w.r.t. x:
sin (log x)
Integrate the following functions w.r.t. x : `log(1 + x)^((1 + x)`
Choose the correct options from the given alternatives :
`int (sin^m x)/(cos^(m+2)x)*dx` =
Choose the correct options from the given alternatives :
`int (x- sinx)/(1 - cosx)*dx` =
Choose the correct options from the given alternatives :
`int sin (log x)*dx` =
Integrate the following with respect to the respective variable : `(sin^6θ + cos^6θ)/(sin^2θ*cos^2θ)`
Integrate the following with respect to the respective variable : cos 3x cos 2x cos x
Integrate the following w.r.t.x : cot–1 (1 – x + x2)
Integrate the following w.r.t.x : `sqrt(x)sec(x^(3/2))*tan(x^(3/2))`
Integrate the following w.r.t.x : e2x sin x cos x
Evaluate the following.
∫ x log x dx
Evaluate the following.
`int "e"^"x" "x"/("x + 1")^2` dx
Evaluate the following.
`int "e"^"x" "x - 1"/("x + 1")^3` dx
Evaluate the following.
`int (log "x")/(1 + log "x")^2` dx
`int ("x" + 1/"x")^3 "dx"` = ______
Evaluate: `int e^x/sqrt(e^(2x) + 4e^x + 13)` dx
Evaluate: `int "dx"/("x"[(log "x")^2 + 4 log "x" - 1])`
`int sin4x cos3x "d"x`
Choose the correct alternative:
`intx^(2)3^(x^3) "d"x` =
`int ("d"x)/(x - x^2)` = ______
`int_0^"a" sqrt("x"/("a" - "x")) "dx"` = ____________.
Evaluate the following:
`int ((cos 5x + cos 4x))/(1 - 2 cos 3x) "d"x`
Evaluate the following:
`int_0^1 x log(1 + 2x) "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.
The integral `int x cos^-1 ((1 - x^2)/(1 + x^2))dx (x > 0)` is equal to ______.
`int e^x [(2 + sin 2x)/(1 + cos 2x)]dx` = ______.
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`.
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
Evaluate:
`intcos^-1(sqrt(x))dx`
Evaluate:
`int((1 + sinx)/(1 + cosx))e^x dx`
Evaluate:
`int e^(logcosx)dx`
Evaluate:
`int (sin(x - a))/(sin(x + a))dx`
Complete the following activity:
`int_0^2 dx/(4 + x - x^2) `
= `int_0^2 dx/(-x^2 + square + square)`
= `int_0^2 dx/(-x^2 + x + 1/4 - square + 4)`
= `int_0^2 dx/ ((x- 1/2)^2 - (square)^2)`
= `1/sqrt17 log((20 + 4sqrt17)/(20 - 4sqrt17))`
