Advertisements
Advertisements
प्रश्न
Evaluate the following : `int x.cos^3x.dx`
Advertisements
उत्तर
cos 3x = 4 cos3x – 3cos x
∴ cos 3x + 3 cos x = 4 cos3x
∴ `int cos^3x = (1)/(4) cos3x + (3)/(4) cosx`
∴ `int cos^3x.dx = (1)/(4) int cos3x.dx + (3)/(4) int cos x.dx`
= `(1)/(4)((sin3x)/3) + (3)/(4) sinx`
= `(sin3x)/(12) + (3sinx)/(4)` ...(1)
Let I = `int x cos^3x.dx`
= `x int cos^3x.dx - int[{d/dx (x) int cos^3x.dx}].dx`
= `x[(sin3x)/(12) + (3sinx)/(4)]- int 1.((sin3x)/(12) + (3sinx)/4).dx` ...[By (1)]
= `(xsin3x)/(12) + (3x sinx)/(4) - (1)/(12) int sin 3x.dx - 3/4 int sin x.dx`
= `(x sin3x)/(12) + (3xsinx)/(4) - (1)/(12) ((-cos3x)/3) - (3)/(4) (- cos x) + c`
= `(1)/(4)[x/3 sin 3x + 1/9 cos3x + 3x sin x + 3 cos x] + c`.
APPEARS IN
संबंधित प्रश्न
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`
`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 log 2x.
Integrate the function in x cos-1 x.
Integrate the function in `(x cos^(-1) x)/sqrt(1-x^2)`.
Integrate the function in `e^x (1 + sin x)/(1+cos x)`.
Integrate the function in `((x- 3)e^x)/(x - 1)^3`.
Find :
`∫(log x)^2 dx`
Evaluate the following:
`int sec^3x.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 : `e^x/x [x (logx)^2 + 2 (logx)]`
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)/(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 : `(3 - 2sinx)/(cos^2x)`
Integrate the following w.r.t.x : cot–1 (1 – x + x2)
Integrate the following w.r.t.x : `log (1 + cosx) - xtan(x/2)`
Integrate the following w.r.t.x : log (log x)+(log x)–2
Integrate the following w.r.t.x : `(1)/(x^3 sqrt(x^2 - 1)`
Integrate the following w.r.t.x : log (x2 + 1)
Integrate the following w.r.t.x : sec4x cosec2x
Solve the following differential equation.
(x2 − yx2 ) dy + (y2 + xy2) dx = 0
Evaluate the following.
`int x^2 e^4x`dx
`int (sinx)/(1 + sin x) "d"x`
`int (sin(x - "a"))/(cos (x + "b")) "d"x`
`int (cos2x)/(sin^2x cos^2x) "d"x`
`int sqrt(tanx) + sqrt(cotx) "d"x`
Choose the correct alternative:
`int ("d"x)/((x - 8)(x + 7))` =
`int tan^-1 sqrt(x) "d"x` is equal to ______.
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.
`int x/((x + 2)(x + 3)) dx` = ______ + `int 3/(x + 3) dx`
`int((4e^x - 25)/(2e^x - 5))dx = Ax + B log(2e^x - 5) + c`, then ______.
Evaluate :
`int(4x - 6)/(x^2 - 3x + 5)^(3/2) 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
`int logx dx = x(1+logx)+c`
Evaluate `int(1 + x + (x^2)/(2!))dx`
Evaluate:
`int((1 + sinx)/(1 + cosx))e^x dx`
Evaluate `int tan^-1x dx`
Evaluate the following.
`intx^3e^(x^2) dx`
If ∫(cot x – cosec2 x)ex dx = ex f(x) + c then f(x) will be ______.
Evaluate the following.
`intx^3/(sqrt(1 + x^4))dx`
Evaluate.
`int(5x^2 - 6x + 3)/(2x - 3) dx`
`∫ sin^(−1)` xdx is equal to ______.
