Advertisements
Advertisements
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`
Advertisements
उत्तर
Let I = `int sqrt(x^2 - a^2)dx`
I = `int sqrt(x^2 - a^2)*1dx`
I = `sqrt(x^2 - a^2)*int1dx - int[d/dx(sqrt(x^2 - a^2))*int1dx]dx`
I = `sqrt(x^2 - a^2)*x - int[1/(2sqrt(x^2 - a^2))*d/dx(x^2 - a^2)*x]dx`
I = `sqrt(x^2 - a^2)*x - int1/(2sqrt(x^2 - a^2))(2x - 0)*x dx`
I = `sqrt(x^2 - a^2)*x - intx/sqrt(x^2 - a^2)*x dx`
I = `xsqrt(x^2 - a^2) - int(x^2 - a^2 + a^2)/(sqrt(x^2 - a^2))dx`
I = `xsqrt(x^2 - a^2) - intsqrt(x^2 - a^2) dx - a^2intdx/(sqrt(x^2 - a^2)`
I = `xsqrt(x^2 - a^2) - I - a^2log|x + sqrt(x^2 - a^2)| + c_1`
∴ 2I = `xsqrt(x^2 - a^2) - a^2log|x + sqrt(x^2 - a^2)| + c_1`
∴ I = `x/2sqrt(x^2-a^2) - a^2/2log|x + sqrt(x^2 - a^2)| + c_1/2`
∴ `intsqrt(x^2 - a^2) dx = x/2sqrt(x^2 - a^2) - a^2/2log|x + sqrt(x^2 - a^2)| + c, "where" c = c_1/2`
APPEARS IN
संबंधित प्रश्न
Integrate the function in x sin x.
Integrate the function in `x^2e^x`.
Integrate the function in (sin-1x)2.
Prove that:
`int sqrt(x^2 + a^2)dx = x/2 sqrt(x^2 + a^2) + a^2/2 log |x + sqrt(x^2 + a^2)| + c`
Evaluate the following : `int (t.sin^-1 t)/sqrt(1 - t^2).dt`
Integrate the following functions w.r.t. x : `sqrt((x - 3)(7 - x)`
Integrate the following functions w.r.t. x : `sqrt(4^x(4^x + 4))`
Integrate the following functions w.r.t. x : `(x + 1) sqrt(2x^2 + 3)`
Integrate the following functions w.r.t. x : `((1 + sin x)/(1 + cos x)).e^x`
Choose the correct options from the given alternatives :
`int (1)/(x + x^5)*dx` = f(x) + c, then `int x^4/(x + x^5)*dx` =
Choose the correct options from the given alternatives :
`int (log (3x))/(xlog (9x))*dx` =
Choose the correct options from the given alternatives :
`int cos -(3)/(7)x*sin -(11)/(7)x*dx` =
Integrate the following w.r.t.x : e2x sin x cos x
Evaluate the following.
`int x^2 e^4x`dx
Evaluate the following.
`int x^2 *e^(3x)`dx
`int (sin(x - "a"))/(cos (x + "b")) "d"x`
`int ("d"x)/(x - x^2)` = ______
`int logx/(1 + logx)^2 "d"x`
`int 1/sqrt(x^2 - 8x - 20) "d"x`
`int "e"^x [x (log x)^2 + 2 log x] "dx"` = ______.
Evaluate the following:
`int_0^1 x log(1 + 2x) "d"x`
`int "dx"/(sin(x - "a")sin(x - "b"))` 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.
Solve: `int sqrt(4x^2 + 5)dx`
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` = ______.
Find: `int e^(x^2) (x^5 + 2x^3)dx`.
`int1/sqrt(x^2 - a^2) dx` = ______
Solution of the equation `xdy/dx=y log y` is ______
`int1/(x+sqrt(x)) 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
`inte^(xloga).e^x dx` is ______
The integrating factor of `ylogy.dx/dy+x-logy=0` is ______.
Evaluate the following.
`int (x^3)/(sqrt(1 + x^4))dx`
Evaluate:
`int e^(ax)*cos(bx + c)dx`
Prove that `int sqrt(x^2 - a^2)dx = x/2 sqrt(x^2 - a^2) - a^2/2 log(x + sqrt(x^2 - a^2)) + c`
Evaluate the following.
`intx^3 e^(x^2)dx`
