Advertisements
Advertisements
प्रश्न
Integrate the following with respect to the respective variable : cos 3x cos 2x cos x
Advertisements
उत्तर
Let I = `int cos 3x cos 2x cos x *dx`
Consider cos 3x cos 2x cos x = `(1)/(2) cos 3x [2 cos 2x cos x]`
= `(1)/(2)cos3x [cos(2x + x) + cos(2x - x)]`
= `(1)/(2)[cos^2 3x + cos3x cosx]`
= `(1)/(4)[2cos^2 3x + 2cos 3x cosx]`
= `(1)/(4)[1 + cos6x + cos(3x + x) + cos(3x - x)]`
= `(1)/(4)[1 + cos6x + cos4x + cos2x]`
∴ I = `(1)/(4) int[1 + cos6x + cos4x + cos2x]*dx`
= `(1)/(4) int 1*dx + 1/4 int cos6x*dx + 1/4 int cos4x*dx + 1/4 int cos2x*dx`
= `x/(4) + (1)/(4)((sin6x)/6) + 1/4((sin4x)/4) + 1/4((sin2x)/2) + c`
= `(1)/(48)[12x + 2sin 6x + 3sin 4x + 6sin2x] + c`.
APPEARS IN
संबंधित प्रश्न
Integrate the function in `(x cos^(-1) x)/sqrt(1-x^2)`.
Integrate the function in x sec2 x.
Integrate the function in tan-1 x.
Integrate the function in x (log x)2.
Integrate the function in e2x sin x.
`int e^x sec x (1 + tan x) dx` equals:
Find :
`∫(log x)^2 dx`
Evaluate the following:
`int x tan^-1 x . dx`
Evaluate the following : `int x^2*cos^-1 x*dx`
Integrate the following functions w.r.t. x : `xsqrt(5 - 4x - x^2)`
Integrate the following functions w.r.t. x : [2 + cot x – cosec2x]ex
Integrate the following functions w.r.t. x : `e^x .(1/x - 1/x^2)`
Integrate the following functions w.r.t. x : `log(1 + x)^((1 + x)`
Integrate the following with respect to the respective variable : `t^3/(t + 1)^2`
Integrate the following w.r.t. x: `(1 + log x)^2/x`
Integrate the following w.r.t.x : log (x2 + 1)
Evaluate the following.
`int "e"^"x" "x"/("x + 1")^2` dx
Choose the correct alternative from the following.
`int (1 - "x")^(-2) "dx"` =
Choose the correct alternative from the following.
`int (("x"^3 + 3"x"^2 + 3"x" + 1))/("x + 1")^5 "dx"` =
Evaluate: `int "dx"/("9x"^2 - 25)`
Evaluate: `int "dx"/(25"x" - "x"(log "x")^2)`
Evaluate: `int "e"^"x"/(4"e"^"2x" -1)` dx
`int (sinx)/(1 + sin x) "d"x`
`int 1/x "d"x` = ______ + c
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"` = ____________.
Find `int_0^1 x(tan^-1x) "d"x`
The value of `int_(- pi/2)^(pi/2) (x^3 + x cos x + tan^5x + 1) dx` is
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`
Find the general solution of the differential equation: `e^((dy)/(dx)) = x^2`.
The integral `int x cos^-1 ((1 - x^2)/(1 + x^2))dx (x > 0)` is equal to ______.
`int_0^1 x tan^-1 x 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`.
`int1/sqrt(x^2 - a^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`
`int(xe^x)/((1+x)^2) dx` = ______
`int (sin^-1 sqrt(x) + cos^-1 sqrt(x))dx` = ______.
The value of `int e^x((1 + sinx)/(1 + cosx))dx` is ______.
Evaluate:
`int1/(x^2 + 25)dx`
Evaluate `int (1 + x + x^2/(2!))dx`
Evaluate the following.
`int x^3 e^(x^2) dx`
Evaluate the following.
`intx^2e^(4x)dx`
Evaluate `int(1 + x + x^2/(2!))dx`.
`∫ sin^(−1)` xdx is equal to ______.
