Advertisements
Advertisements
Question
`int 1/sqrt(2x^2 - 5) "d"x`
Advertisements
Solution
Let I = `int 1/sqrt(2x^2 - 5) "d"x`
= `int 1/sqrt(2(x^2 - 5/2)) "d"x`
= `1/sqrt(2) int 1/sqrt(x^2 - (sqrt(5)/sqrt(2))^2) "d"x`
= `1/sqrt(2) log|x + sqrt(x^2 - (sqrt(5)/sqrt(2))^2)| + "c"`
∴ I = `1/sqrt(2) log|x + sqrt(x^2 - 5/2)| + "c"`
APPEARS IN
RELATED QUESTIONS
Prove that: `int sqrt(a^2 - x^2) * dx = x/2 * sqrt(a^2 - x^2) + a^2/2 * sin^-1(x/a) + c`
Integrate : sec3 x w. r. t. x.
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
`int1/xlogxdx=...............`
(A)log(log x)+ c
(B) 1/2 (logx )2+c
(C) 2log x + c
(D) log x + c
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 `e^x (1 + sin x)/(1+cos x)`.
Integrate the function in `e^x (1/x - 1/x^2)`.
Evaluate the following:
`int x^2 sin 3x dx`
Evaluate the following : `int x.sin^2x.dx`
Evaluate the following: `int x.sin^-1 x.dx`
Evaluate the following : `int (t.sin^-1 t)/sqrt(1 - t^2).dt`
Evaluate the following : `int cos sqrt(x).dx`
Evaluate the following : `int x.cos^3x.dx`
Integrate the following functions w.r.t. x : `sqrt((x - 3)(7 - x)`
Integrate the following functions w.r.t.x:
`e^(5x).[(5x.logx + 1)/x]`
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 : cos 3x cos 2x cos x
Evaluate the following.
`int e^x (1/x - 1/x^2)`dx
Evaluate: `int "dx"/sqrt(4"x"^2 - 5)`
`int ["cosec"(logx)][1 - cot(logx)] "d"x`
`int sin4x cos3x "d"x`
Choose the correct alternative:
`intx^(2)3^(x^3) "d"x` =
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
Evaluate `int 1/(x(x - 1)) "d"x`
`int [(log x - 1)/(1 + (log x)^2)]^2`dx = ?
`int log x * [log ("e"x)]^-2` dx = ?
`int tan^-1 sqrt(x) "d"x` is equal to ______.
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: `int e^x.sin2xdx`
If `int(2e^(5x) + e^(4x) - 4e^(3x) + 4e^(2x) + 2e^x)/((e^(2x) + 4)(e^(2x) - 1)^2)dx = tan^-1(e^x/a) - 1/(b(e^(2x) - 1)) + C`, where C is constant of integration, then value of a + b is equal to ______.
`int_0^1 x tan^-1 x dx` = ______.
Find `int (sin^-1x)/(1 - x^2)^(3//2) dx`.
`int(1-x)^-2 dx` = ______
`int1/sqrt(x^2 - a^2) dx` = ______
Evaluate the following.
`int x^3 e^(x^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(xe^x)/((1+x)^2) dx` = ______
Evaluate `int(1 + x + (x^2)/(2!))dx`
Evaluate:
`int e^(ax)*cos(bx + c)dx`
Evaluate:
`inte^x sinx dx`
Evaluate:
`int e^(logcosx)dx`
If u and v are two differentiable functions of x, then prove that `intu*v*dx = u*intv dx - int(d/dx u)(intv dx)dx`. Hence evaluate: `intx cos x dx`
Evaluate the following.
`int x sqrt(1 + x^2) dx`
Evaluate the following.
`intx^2e^(4x)dx`
Evaluate:
`inte^x "cosec" x(1 - cot x)dx`
