Advertisements
Advertisements
प्रश्न
`int tan^-1 sqrt(x) "d"x` is equal to ______.
विकल्प
`(x + 1) tan^-1 sqrt(x) - sqrt(x) + "C"`
`x tan^-1 sqrt(x) - sqrt(x) + "C"`
`sqrt(x) - x tan^-1 sqrt(x) + "C"`
`sqrt(x) - (x + 1) tan^-1 sqrt(x) + "C"`
Advertisements
उत्तर
`int tan^-1 sqrt(x) "d"x` is equal to `(x + 1) tan^-1 sqrt(x) - sqrt(x) + "C"`.
Explanation:
Let I = `int 1 * tan^-1 sqrt(x) "d"x`
= `tan^-1 sqrt(x) int 1 "d"x - int[(tan^-1 sqrt(x))"'" int 1"d"x]"d"x`
= `tan^-1 sqrt(x) * x - int 1/(1 + x) * 1/(2sqrt(x)) * x"d"x` ....[Integrating by parrts]
= `xtan^-1 sqrt(x) - 1/2 int sqrt(x)/(1 + x) "d"x`
Put x = t2
⇒ dx = 2t dt
∴ I = `xtan^-1 sqrt(x) - int "t"^2/(1 + "t"^2) "d"x`
= `xtan^-1 sqrt(x) - int (1 - 1/(1 + "t"^2))"dt"`
= `xtan^-1 sqrt(x) - "t" + tan^-1 1 + "C"`
= `xtan^-1 sqrt(x) - sqrt(x) + tan^-1 sqrt(x) + "C"`
= `(x + 1) tan^-1 sqrt(x) - sqrt(x) + "C"`
APPEARS IN
संबंधित 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`
Integrate the function in x sin 3x.
Integrate the function in `x^2e^x`.
Integrate the function in x sec2 x.
Integrate the function in x (log x)2.
Integrate the function in `e^x (1 + sin x)/(1+cos x)`.
Evaluate the following:
`int x^2 sin 3x dx`
Evaluate the following : `int e^(2x).cos 3x.dx`
Evaluate the following : `int x^2*cos^-1 x*dx`
Evaluate the following : `int sin θ.log (cos θ).dθ`
Evaluate the following : `int cos(root(3)(x)).dx`
Integrate the following functions w.r.t. x : cosec (log x)[1 – cot (log x)]
Choose the correct options from the given alternatives :
`int tan(sin^-1 x)*dx` =
Choose the correct options from the given alternatives :
`int cos -(3)/(7)x*sin -(11)/(7)x*dx` =
Evaluate the following.
`int "e"^"x" "x - 1"/("x + 1")^3` dx
Evaluate the following.
`int "e"^"x" [(log "x")^2 + (2 log "x")/"x"]` dx
Evaluate the following.
`int [1/(log "x") - 1/(log "x")^2]` dx
Evaluate: `int "dx"/("x"[(log "x")^2 + 4 log "x" - 1])`
Evaluate: `int "dx"/(5 - 16"x"^2)`
`int ("d"x)/(x - x^2)` = ______
Evaluate the following:
`int_0^1 x log(1 + 2x) "d"x`
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 the general solution of the differential equation: `e^((dy)/(dx)) = x^2`.
If `π/2` < x < π, then `intxsqrt((1 + cos2x)/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 logx dx = x(1+logx)+c`
Solve the following
`int_0^1 e^(x^2) x^3 dx`
Evaluate:
`intcos^-1(sqrt(x))dx`
`int (sin^-1 sqrt(x) + cos^-1 sqrt(x))dx` = ______.
Evaluate `int tan^-1x dx`
Complete the following activity:
`int_0^2 dx/(4 + x - x^2) `
= `int_0^2 dx/(-x^2 + square + square)`
= `int_0^2 dx/(-x^2 + x + 1/4 - square + 4)`
= `int_0^2 dx/ ((x- 1/2)^2 - (square)^2)`
= `1/sqrt17 log((20 + 4sqrt17)/(20 - 4sqrt17))`
Evaluate the following.
`intx^3 e^(x^2) dx`
The value of `inta^x.e^x dx` equals
The value of `int (x sin^-1)/(sqrt(1 - x^2)) dx` is equal to:
If \(u\) and \(v\) are differentiable functions, which formula is used for integration by parts?
Evaluate \[\int \frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx.\]
Evaluate \[\int x\cos x\,dx.\]
