Advertisements
Advertisements
Question
Prove that: `int sqrt(a^2 - x^2) * dx = x/2 * sqrt(a^2 - x^2) + a^2/2 * sin^-1(x/a) + c`
Advertisements
Solution
Let I = `int sqrt(a^2 - x^2) dx`
= `int sqrt(a^2 - x^2)*1 dx`
= `sqrt(a^2 - x^2)* int 1 dx - int [d/dx (sqrt(a^2 - x^2))* int 1 dx]dx`
= `sqrt(a^2 - x^2)*x - int [1/(2sqrt(a^2 - x^2))*d/dx (a^2 - x^2)*x]dx`
= `sqrt(a^2 - x^2)*x - int 1/(2sqrt(a^2 - x^2))(0 - 2x)*x dx`
= `sqrt(a^2 - x^2)*x - int (-x)/sqrt(a^2 - x^2)*x dx`
= `xsqrt(a^2 - x^2) - int (a^2 - x^2 - a^2)/sqrt(a^2 - x^2)dx`
= `xsqrt(a^2 - x^2) - int sqrt(a^2 - x^2)dx + a^2 int dx/sqrt(a^2 - x^2)`
= `xsqrt(a^2 - x^2) - I + a^2sin^-1(x/a) + c_1`
∴ 2I = `xsqrt(a^2 - x^2) + a^2sin^-1(x/a) + c_1`
∴ I = `x/2 sqrt(a^2 - x^2) + a^2/2 sin^-1(x/a) + c_1/2`
∴ `int sqrt(a^2 - x^2)dx = x/2 sqrt(a^2 - x^2) + a^2/2sin^-1(x/a) + c`, where `c = c_1/2`.
RELATED QUESTIONS
Integrate the function in x2 log x.
Integrate the function in x cos-1 x.
Evaluate the following : `int x^2tan^-1x.dx`
Evaluate the following : `int x^3.logx.dx`
Evaluate the following : `int x^2*cos^-1 x*dx`
Evaluate the following:
`int x.sin 2x. cos 5x.dx`
Integrate the following functions w.r.t. x : `sec^2x.sqrt(tan^2x + tan x - 7)`
Integrate the following functions w.r.t. x : `((1 + sin x)/(1 + cos x)).e^x`
Integrate the following functions w.r.t. x : `e^x/x [x (logx)^2 + 2 (logx)]`
Integrate the following functions w.r.t.x:
`e^(5x).[(5x.logx + 1)/x]`
Choose the correct options from the given alternatives :
`int (log (3x))/(xlog (9x))*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 : `(3 - 2sinx)/(cos^2x)`
Integrate the following w.r.t.x : log (x2 + 1)
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 "e"^"x" "x"/("x + 1")^2` dx
Evaluate: Find the primitive of `1/(1 + "e"^"x")`
Evaluate: `int "dx"/(3 - 2"x" - "x"^2)`
Evaluate: `int "dx"/(25"x" - "x"(log "x")^2)`
`int 1/x "d"x` = ______ + c
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 log x) "d"x`
Evaluate `int (2x + 1)/((x + 1)(x - 2)) "d"x`
`int cot "x".log [log (sin "x")] "dx"` = ____________.
Find: `int e^x.sin2xdx`
Find the general solution of the differential equation: `e^((dy)/(dx)) = x^2`.
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 ______.
Find `int (sin^-1x)/(1 - x^2)^(3//2) dx`.
Find `int e^x ((1 - sinx)/(1 - cosx))dx`.
Evaluate:
`int(1+logx)/(x(3+logx)(2+3logx)) dx`
`int(3x^2)/sqrt(1+x^3) dx = sqrt(1+x^3)+c`
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` = ______
`int(f'(x))/sqrt(f(x)) dx = 2sqrt(f(x))+c`
Evaluate:
`int((1 + sinx)/(1 + cosx))e^x dx`
The value of `int e^x((1 + sinx)/(1 + cosx))dx` is ______.
Evaluate the following.
`intx^3 e^(x^2) dx`
Evaluate:
`int1/(x^2 + 25)dx`
Evaluate the following.
`intx^3e^(x^2) dx`
Evaluate:
`int x^2 cos x dx`
Evaluate:
`inte^x "cosec" x(1 - cot x)dx`
Evaluate the following.
`intx^3/(sqrt(1 + x^4))dx`
`∫ sin^(−1)` xdx is equal to ______.
