Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int x.sin 2x. cos 5x.dx`
Advertisements
उत्तर
Let I = `int x.sin 2x. cos 5x.dx`
`sin 2x cos 5x = (1)/(2)[2 sin2x cos5x]`
= `(1)/(2)[sin(2x+ 5x) + sin(2x - 5x)]`
= `(1)/(2)[sin7x - sin3x]`
∴ `int sin 2x cos 5x .dx = (1)/(2)[int sin 7x ..dx - intsin 3x.dx]`
= `(1)/(2)((-cos7x)/7) - (1)/(2) ((- cos3x)/3)`
= `-(1)/(14) cos7x + (1)/(6) cos3x` ...(1)
I = `int x sin 2x cos 5x.dx`
= `x int sin 2x cos 5x.dx - int [d/dx (x) int sin 2x cos 5x.dx].dx`
= `x[-1/14 cos7x + 1/6 cos 3x] - int 1.(-1/14 cos7x + 1/6 cos3x).dx` ...[By (1)]
= `-x/(14) cos7x + x/(6) cos3x + (1)/(14) int cos7x.dx - (1)/(6) int cos 3x.dx`
= `-x/(14) cos7x + x/(6) cos3x + (1)/(14) ((sin7x)/7) - (1)/(6) ((sin3x)/3) + c`
= `- x/(14) cos7x + x/(6) cos3x + (sin7x)/(98) - (sin3x)/(18) + c`.
APPEARS IN
संबंधित प्रश्न
Integrate : sec3 x w. r. t. x.
Integrate the function in `x^2e^x`.
Integrate the function in x tan-1 x.
Integrate the function in x (log x)2.
Integrate the function in `(xe^x)/(1+x)^2`.
Integrate the function in `sin^(-1) ((2x)/(1+x^2))`.
`int e^x sec x (1 + tan x) dx` equals:
Evaluate the following : `int x^2.log x.dx`
Evaluate the following : `int log(logx)/x.dx`
Evaluate the following : `int x.cos^3x.dx`
Integrate the following functions w.r.t. x : `sqrt(4^x(4^x + 4))`
Integrate the following functions w.r.t. x : `xsqrt(5 - 4x - x^2)`
Integrate the following functions w.r.t. x : `e^x/x [x (logx)^2 + 2 (logx)]`
Choose the correct options from the given alternatives :
`int sin (log x)*dx` =
Integrate the following w.r.t.x : e2x sin x cos x
Evaluate the following.
∫ x log x dx
Evaluate the following.
`int x^2 e^4x`dx
Evaluate the following.
`int "e"^"x" "x - 1"/("x + 1")^3` dx
`int ("x" + 1/"x")^3 "dx"` = ______
Choose the correct alternative from the following.
`int (("e"^"2x" + "e"^"-2x")/"e"^"x") "dx"` =
Choose the correct alternative from the following.
`int (("x"^3 + 3"x"^2 + 3"x" + 1))/("x + 1")^5 "dx"` =
Evaluate: `int ("ae"^("x") + "be"^(-"x"))/("ae"^("x") - "be"^(−"x"))` dx
Evaluate: `int "dx"/(5 - 16"x"^2)`
Evaluate: `int "e"^"x"/(4"e"^"2x" -1)` dx
`int 1/(4x + 5x^(-11)) "d"x`
`int ["cosec"(logx)][1 - cot(logx)] "d"x`
Choose the correct alternative:
`int ("d"x)/((x - 8)(x + 7))` =
`int (x^2 + x - 6)/((x - 2)(x - 1)) "d"x` = x + ______ + c
`int logx/(1 + logx)^2 "d"x`
`int [(log x - 1)/(1 + (log x)^2)]^2`dx = ?
`int_0^"a" sqrt("x"/("a" - "x")) "dx"` = ____________.
`int "dx"/(sin(x - "a")sin(x - "b"))` is equal to ______.
The value of `int_(- pi/2)^(pi/2) (x^3 + x cos x + tan^5x + 1) dx` is
`int x/((x + 2)(x + 3)) dx` = ______ + `int 3/(x + 3) dx`
Find: `int e^x.sin2xdx`
Find the general solution of the differential equation: `e^((dy)/(dx)) = x^2`.
`int_0^1 x tan^-1 x dx` = ______.
`int((4e^x - 25)/(2e^x - 5))dx = Ax + B log(2e^x - 5) + c`, then ______.
`int1/sqrt(x^2 - a^2) dx` = ______
Evaluate:
`int(1+logx)/(x(3+logx)(2+3logx)) dx`
`int logx dx = x(1+logx)+c`
Evaluate:
`int((1 + sinx)/(1 + cosx))e^x dx`
Evaluate:
`int e^(ax)*cos(bx + c)dx`
`int (sin^-1 sqrt(x) + cos^-1 sqrt(x))dx` = ______.
Evaluate the following.
`intx^3/sqrt(1+x^4) dx`
Evaluate the following.
`intx^3/(sqrt(1 + x^4))dx`
Evaluate.
`int(5x^2 - 6x + 3)/(2x - 3) dx`
