Advertisements
Advertisements
प्रश्न
Integrate the following w.r.t.x : `log (1 + cosx) - xtan(x/2)`
Advertisements
उत्तर
Let I = `int [log(1 + cosx) - xtan(x/2)]*dx`
= `int [log(1 + cos.x)*1dx - intxtan (x/2)*dx`
= `[log(1 + cosx)]* int 1dx - int {d/dx [log (1 + cosx)]* int 1dx}*dx - xtan (x/2)*dx`
= `[log (1 + cosx)]*(x) - int 1/(1 + cosx)*(0 - sin x)*xdx - int x tan (x/2)*dx`
= `x*log(1 + cosx) + intx* (sinx)/(1 + cosx)*dx - int xtan (x/2)*dx + c`
= `x*log(1 + cosx) + intx*(2sin(x/2)*cos(x/2))/(2cos^2(x/2)*dx - int xtan (x/2)*dx + c`
= `xlog (1 + cosx) + int x*tan(x/2)*dx - intxtan(x/2)*dx + c`
= x·log(1 + cosx) + 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 cos^(-1) x)/sqrt(1-x^2)`.
Integrate the function in (x2 + 1) log x.
Integrate the function in `((x- 3)e^x)/(x - 1)^3`.
`intx^2 e^(x^3) dx` equals:
Prove that:
`int sqrt(x^2 + a^2)dx = x/2 sqrt(x^2 + a^2) + a^2/2 log |x + sqrt(x^2 + a^2)| + c`
Evaluate the following : `int x^2tan^-1x.dx`
Evaluate the following : `int x^3.logx.dx`
Evaluate the following: `int x.sin^-1 x.dx`
Evaluate the following : `int sin θ.log (cos θ).dθ`
Evaluate the following : `int(sin(logx)^2)/x.log.x.dx`
Integrate the following functions w.r.t. x : `x^2 .sqrt(a^2 - x^6)`
Integrate the following functions w.r.t. x : `sqrt((x - 3)(7 - x)`
Integrate the following functions w.r.t. x : cosec (log x)[1 – cot (log x)]
Choose the correct options from the given alternatives :
`int (1)/(x + x^5)*dx` = f(x) + c, then `int x^4/(x + x^5)*dx` =
Choose the correct options from the given alternatives :
`int (log (3x))/(xlog (9x))*dx` =
Integrate the following with respect to the respective variable : `(sin^6θ + cos^6θ)/(sin^2θ*cos^2θ)`
Integrate the following w.r.t. x: `(1 + log x)^2/x`
Integrate the following w.r.t.x : `(1)/(xsin^2(logx)`
Integrate the following w.r.t.x : log (log x)+(log x)–2
Evaluate the following.
`int "e"^"x" "x"/("x + 1")^2` dx
Evaluate the following.
`int [1/(log "x") - 1/(log "x")^2]` dx
Evaluate the following.
`int (log "x")/(1 + log "x")^2` dx
Evaluate: Find the primitive of `1/(1 + "e"^"x")`
Evaluate: `int "dx"/("x"[(log "x")^2 + 4 log "x" - 1])`
Evaluate: `int "dx"/(5 - 16"x"^2)`
Evaluate:
∫ (log x)2 dx
`int (sin(x - "a"))/(cos (x + "b")) "d"x`
`int 1/sqrt(2x^2 - 5) "d"x`
`int sin4x cos3x "d"x`
`int"e"^(4x - 3) "d"x` = ______ + c
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
`int x/((x + 2)(x + 3)) dx` = ______ + `int 3/(x + 3) dx`
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` = ______.
Find `int e^x ((1 - sinx)/(1 - cosx))dx`.
Evaluate :
`int(4x - 6)/(x^2 - 3x + 5)^(3/2) 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
Evaluate `int(1 + x + (x^2)/(2!))dx`
Evaluate:
`int((1 + sinx)/(1 + cosx))e^x dx`
Evaluate the following.
`int x^3 e^(x^2) dx`
Evaluate the following.
`intx^3/sqrt(1+x^4)`dx
The value of `inta^x.e^x dx` equals
Evaluate the following.
`intx^3/(sqrt(1 + x^4))dx`
Evaluate.
`int(5x^2 - 6x + 3)/(2x - 3) dx`
The value of `int (x sin^-1)/(sqrt(1 - x^2)) dx` is equal to:
