Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int sec^3x.dx`
Advertisements
उत्तर
Let I = `int sec^3x.dx`
= `int sec x sec^2x.dx`
= `sec x int sec^2x.dx - int[d/dx(secx) int sec^2x.dx].dx`
= `secx tanx- int (secx tanx)(tanx).dx`
= `secx tanx - int secx tan^2x.dx`
= `secx tanx - int secx (sec^2x - 1).dx`
= `secx tanx - int sec^3x.dx + int secx.dx`
∴ I = sec x tan x – I + log |sec x + tanx|
∴ 2I = sec x tan x + log |sec x + tan x|
∴ I = `(1)/(2)[secx tanx + log |secx + tan|] + 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
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 x2 log x.
Integrate the function in `(x cos^(-1) x)/sqrt(1-x^2)`.
Integrate the function in `(xe^x)/(1+x)^2`.
Evaluate the following : `int x^3.tan^-1x.dx`
Evaluate the following : `int x^3.logx.dx`
Evaluate the following : `int e^(2x).cos 3x.dx`
Evaluate the following: `int x.sin^-1 x.dx`
Evaluate the following : `int x^2*cos^-1 x*dx`
Evaluate the following : `int cos(root(3)(x)).dx`
Integrate the following functions w.r.t. x : `e^(2x).sin3x`
Integrate the following functions w.r.t. x : `x^2 .sqrt(a^2 - x^6)`
Integrate the following functions w.r.t. x : `log(1 + x)^((1 + x)`
Choose the correct options from the given alternatives :
`int tan(sin^-1 x)*dx` =
If f(x) = `sin^-1x/sqrt(1 - x^2), "g"(x) = e^(sin^-1x)`, then `int f(x)*"g"(x)*dx` = ______.
Choose the correct options from the given alternatives :
`int (1)/(cosx - cos^2x)*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: `(1 + log x)^2/x`
Integrate the following w.r.t.x : `sqrt(x)sec(x^(3/2))*tan(x^(3/2))`
Evaluate the following.
`int x^2 e^4x`dx
Choose the correct alternative from the following.
`int (1 - "x")^(-2) "dx"` =
Evaluate: `int "dx"/("x"[(log "x")^2 + 4 log "x" - 1])`
Evaluate: `int "dx"/(25"x" - "x"(log "x")^2)`
Choose the correct alternative:
`int ("d"x)/((x - 8)(x + 7))` =
`int(x + 1/x)^3 dx` = ______.
`int 1/x "d"x` = ______ + c
Evaluate `int 1/(4x^2 - 1) "d"x`
∫ log x · (log x + 2) dx = ?
`int_0^"a" sqrt("x"/("a" - "x")) "dx"` = ____________.
Evaluate the following:
`int_0^pi x log sin x "d"x`
The value of `int_0^(pi/2) log ((4 + 3 sin x)/(4 + 3 cos x)) dx` is
`int 1/sqrt(x^2 - 9) dx` = ______.
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.
Find: `int e^x.sin2xdx`
If `int(x + (cos^-1 3x)^2)/sqrt(1 - 9x^2)dx = 1/α(sqrt(1 - 9x^2) + (cos^-1 3x)^β) + C`, where C is constant of integration , then (α + 3β) is equal to ______.
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` = ______.
`intsqrt(1+x) dx` = ______
Evaluate `int(3x-2)/((x+1)^2(x+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
Solve the following
`int_0^1 e^(x^2) x^3 dx`
Evaluate the following.
`intx^3 e^(x^2) dx`
Evaluate:
`int1/(x^2 + 25)dx`
Evaluate the following.
`intx^3/sqrt(1+x^4)dx`
Evaluate the following.
`intx^3/sqrt(1+x^4) dx`
