Advertisements
Advertisements
Question
Evaluate the following : `int x^2*cos^-1 x*dx`
Advertisements
Solution
Let I = `int x^2.cos^-1 x*dx`
= `int (cos^-1x)*x^2dx`
= `(cos^-1x) int x^2*dx- int d/dx(cos^-1x) int x^2*dx]*dx`
= `(cos^-1x) (x^3/3) - int ((-1)/sqrt(1 - x^2)) (x^3/3)*dx`
= `x^3/(3) cos^-1x + (1)/(3) int (x^2.x)/sqrt(1 - x^2)*dx`
In `int x^3/sqrt(1 - x^2)*dx`, put 1 – x2 = t
∴ – 2x.dx= dt
∴ x.dx = `-(1)/(2)dt`
Also, x2 = 1 – t
∴ I = `x^3/(3) cos^-1x + (1)/(3) int ((1 - t))/sqrt(t) (-1/2)*dt`
= `x^3/(3) cos^-1x - (1)/(6) int (1/sqrt(t) - sqrt(t))*dt`
= `x^3/(3) cos^-1x - (1)/(6) int t^(-1/2) dt + (1)/(6) int t^(1/2)*dt`
= `x^3/(3) cos^-1x - (1)/(6) (t^(1/2)/(1/2)) + (1)/(6) t^(3/2)/(3/2) + c`
= `x^3/(3) cos^-1x - (1)/(3)sqrt(1 - x^2) + (1)/(9)(1 - x^2)^(3/2) + c`.
APPEARS IN
RELATED QUESTIONS
Integrate the function in x sin x.
Integrate the function in x2 log x.
Integrate the function in x sin−1 x.
Integrate the function in e2x sin x.
Integrate the function in `sin^(-1) ((2x)/(1+x^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 x^2tan^-1x.dx`
Evaluate the following : `int e^(2x).cos 3x.dx`
Evaluate the following : `int (t.sin^-1 t)/sqrt(1 - t^2).dt`
Evaluate the following : `int(sin(logx)^2)/x.log.x.dx`
Evaluate the following: `int logx/x.dx`
Evaluate the following : `int cos(root(3)(x)).dx`
Integrate the following functions w.r.t.x:
`e^-x cos2x`
Integrate the following functions w.r.t. x : `x^2 .sqrt(a^2 - x^6)`
Integrate the following functions w.r.t. x : `(x + 1) sqrt(2x^2 + 3)`
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 cos -(3)/(7)x*sin -(11)/(7)x*dx` =
Integrate the following w.r.t. x: `(1 + log x)^2/x`
Integrate the following w.r.t.x : cot–1 (1 – x + x2)
Integrate the following w.r.t.x : e2x sin x cos x
Integrate the following w.r.t.x : sec4x cosec2x
Evaluate the following.
`int x^3 e^(x^2)`dx
Evaluate the following.
`int "e"^"x" [(log "x")^2 + (2 log "x")/"x"]` dx
`int ("x" + 1/"x")^3 "dx"` = ______
Evaluate: `int "dx"/("9x"^2 - 25)`
Evaluate: `int "dx"/(5 - 16"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
`int log x * [log ("e"x)]^-2` dx = ?
Find `int_0^1 x(tan^-1x) "d"x`
Evaluate the following:
`int ((cos 5x + cos 4x))/(1 - 2 cos 3x) "d"x`
`int x/((x + 2)(x + 3)) dx` = ______ + `int 3/(x + 3) dx`
`int 1/sqrt(x^2 - a^2)dx` = ______.
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 ______.
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 ______.
If `π/2` < x < π, then `intxsqrt((1 + cos2x)/2)dx` = ______.
`intsqrt(1+x) 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(f'(x))/sqrt(f(x)) dx = 2sqrt(f(x))+c`
Solve the following
`int_0^1 e^(x^2) x^3 dx`
Evaluate:
`int (sin(x - a))/(sin(x + a))dx`
Evaluate the following.
`intx^3/sqrt(1+x^4)dx`
Evaluate the following.
`intx^3 e^(x^2) dx`
Evaluate `int (1 + x + x^2/(2!))dx`
If f′(x) = 4x3 − 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x).
Evaluate:
`inte^x "cosec" x(1 - cot x)dx`
