Advertisements
Advertisements
प्रश्न
Find `int_0^1 x(tan^-1x) "d"x`
Advertisements
उत्तर
I = `int_0^1x(tan^-1x)^2 "d"x`
Integrating by parts, we have
I = `x^2/2[(tan^-1x)^2]_0^1 - 1/2 int_0^1 x^2 * 2 (tan^-1x)/(1 + x^2) "d"x`
= `pi^2/32 - int_0^1 x^2/(1 + x) * tan^-1 x"d"x`
= `pi^2/32 - 1_1`, where I1 = `int_0^1 x^2/(1 + x^2) tan^-1 x"d"x`
Now I1 = `int_0^1 (x^2 + 1 - 1)/(1 + x^2) tan^-1x "d"x`
= `int_0^1 tan^-1 x"d"x - int_0^1 1/(1 + x^2) tan^-1 x"d"x`
= `"I"_2 - 1/2 ((tan^-1x)^2)_0^1`
= `"I"_2 - pi^2/32`
Here I2 = `int_0^1 tan^-1 x"d"x = (x tan^-1x)_0^1 - int_0^1 x/(1 + x^2) "d"x`
= `pi/4 - 1/2(log|1 + x^2|)_0^1`
= `pi/4 - 1/2 log2`
Thus I2 = `pi/4 - 1/2 log 2 - pi^2/32`
Therefore, I = `pi^2/32 - pi/4 + 1/2 log2 + pi^2/32`
= `pi^2/16 - pi/4 + 1/2 log2`
= `(pi^2 - 4pi)/16 + log sqrt(2)`.
APPEARS IN
संबंधित प्रश्न
Evaluate `int_0^(pi)e^2x.sin(pi/4+x)dx`
Integrate the function in x sin x.
Integrate the function in x tan-1 x.
Integrate the function in `(xe^x)/(1+x)^2`.
Integrate the function in `e^x (1 + sin x)/(1+cos x)`.
`intx^2 e^(x^3) dx` equals:
`int e^x sec x (1 + tan x) dx` equals:
Evaluate the following:
`int x tan^-1 x . dx`
Evaluate the following : `int x^2tan^-1x.dx`
Evaluate the following: `int x.sin^-1 x.dx`
Evaluate the following : `int cos(root(3)(x)).dx`
Integrate the following functions w.r.t. x : `xsqrt(5 - 4x - x^2)`
Integrate the following functions w.r.t. x : `e^x/x [x (logx)^2 + 2 (logx)]`
Integrate the following functions w.r.t. x : `log(1 + x)^((1 + x)`
Choose the correct options from the given alternatives :
`int (log (3x))/(xlog (9x))*dx` =
Integrate the following with respect to the respective variable : `(sin^6θ + cos^6θ)/(sin^2θ*cos^2θ)`
Integrate the following with respect to the respective variable : cos 3x cos 2x cos x
Solve the following differential equation.
(x2 − yx2 ) dy + (y2 + xy2) dx = 0
Evaluate the following.
`int "e"^"x" "x - 1"/("x + 1")^3` dx
Evaluate: Find the primitive of `1/(1 + "e"^"x")`
Evaluate: `int ("ae"^("x") + "be"^(-"x"))/("ae"^("x") - "be"^(−"x"))` dx
Evaluate: `int e^x/sqrt(e^(2x) + 4e^x + 13)` dx
Evaluate: `int "dx"/(5 - 16"x"^2)`
`int (cos2x)/(sin^2x cos^2x) "d"x`
`int sin4x cos3x "d"x`
`int(x + 1/x)^3 dx` = ______.
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 cot "x".log [log (sin "x")] "dx"` = ____________.
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 the general solution of the differential equation: `e^((dy)/(dx)) = x^2`.
`intsqrt(1+x) dx` = ______
`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
`int logx dx = x(1+logx)+c`
Evaluate:
`int (sin(x - a))/(sin(x + a))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:
`int1/(x^2 + 25)dx`
Evaluate:
`int x^2 cos x dx`
Evaluate `int(1 + x + x^2/(2!))dx`.
