integral from (0 to pi/2 ) `int_0^(pi/2) (tanx)^2dx`

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tiburtius | High School Teacher | (Level 2) Educator

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`int_0^(pi/2) (tan x)^2dx=`

You can use the fact that `tan x =(sin x)/(cos x)` to get

`int_0^(pi/2)(sin^2 x)/(cos^2 x)dx=`` `

Now you write numerator as `sin^2 x=1-cos^2x`

`int_0^(pi/2)(1-cos^2x)/(cos^2x)dx=int_0^(pi/2)dx/cos^2x-int(cos^2x)/(cos^2x)dx=tan x|_0^(pi/2)-x|_0^(pi/2)=`

`tan(pi/2)-0-pi/2+0=tan(pi/2)-pi/2`

` ` Now since tangent is not defined for `pi/2`the above expression cannot be evaluated i.e. integral does not converge.

 

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the definite integral such that:

`int_0^(pi/2) (tan x)^2 dx`

You need to use the following substitution, such that:

`tan(x/2) = t => (1/2)(1 + tan^2(x/2))dx = dt`

`(2dt)/(1 + t^2) = dx`

`tan x = tan 2(x/2) = (2tan(x/2))/(1 - tan^2(x/2))`

`tan x = (2t)/(1 - t^2)`

Changing the limits of integration yields:

`x = 0 => tan 0 = 0`

x = => tan (pi/4) = 1

Changing the variable yields:

`int_0^(pi/2) (4t^2)/((1 - t^2)^2) (2dt)/(1 + t^2)`

`8 int_0^(pi/2) (t^2)/((1 - t^2)^2(1 + t^2)) (dt)`

You need to use the partial fraction decomposition, such that:

`(t^2)/((1 - t^2)^2(1 + t^2)) = A/(1 - t) + B/((1 - t)^2) + C/(1 + t) + D/((1 + t)^2) + (Et + F)/(1 + t^2)`

Bringing the fractions to a common denominator yields:

`t^2 = A(1 - t)(1 + t)^2(1 + t^2) + B(1 + t)^2(1 + t^2) + C(1 - t)^2(1 + t)(1 + t^2) + D(1 - t)^2(1 + t^2) + (Et + F)(1 - t^2)^2`

`t^2 = A(1 - t^2)(1+t)(1 + t^2) + B(1 + 2t + t^2)(1 + t^2) + C(1 - t^2)(1 - t)(1 + t^2) + D(1 - 2t + t^2)(1 + t^2) + (Et + F)(1 - 2t^2 + t^4)`

`t^2 = A(1 - t^4)(1 + t) + B + Bt^2 + 2Bt + 2Bt^3 + Bt^2 + Bt^4 + C(1 - t^4)(1 - t) + D + Dt^2 -2Dt - 2Dt^3 + Dt^2 + Dt^4 + Et - 2Et^3 + Et^5 + F - 2Ft^2 + Ft^4`

`t^2 = A + At - At^4 - At^5 + B + Bt^2 + 2Bt + 2Bt^3 + Bt^2 + Bt^4 + C - Ct - Ct^4 + Ct^5 + D + Dt^2 -2Dt - 2Dt^3 + Dt^2 + Dt^4 + Et - 2Et^3 + Et^5 + F - 2Ft^2 + Ft^4`

`t^2 = t^5(-A + C + E) + t^4(-A + B - C + D + F) + t^3(2B - 2D - 2E) + t^2(B + B + D + D - 2F) + t(A + 2B - C - 2D + E) + A + B + C + D + F`

Equating the coefficients of like powers yields:

`{(-A + C + E = 0),(-A + B - C + D + F = 0),(2B - 2D - 2E = 0),(2B + 2D - 2F = 1),(A + 2B - C - 2D + E = 0),(A + B + C + D + F = 0):}`

`C + E + B - C + D + F = 0 => E + B + D + F = 0`

`B - D = E => B - D + B + D + F = 0 => 2B = -F`

`-3F + 2D = 1`

`C + E + 3E - C = 0 => 4E = 0 => E = 0`

`2C + 3E + 1 + 4F = 0 => 2C + 4F = -1`

`B - D = 0 => B = D`

`A = C`

 

`F = -1/2 => B = D = 1/4 A = C = 1/2`

 

`(t^2)/((1 - t^2)^2(1 + t^2)) = 1/(2(1 - t)) + 1/(4(1 - t)^2) + 1/(2(1 + t)) + 1/(4(1 + t)^2) - 1/(2(1 + t^2))`

Integrating both sides, yields:

`int_0^1 (t^2)/((1 - t^2)^2(1 + t^2)) dt = int_0^1 1/(2(1 - t)) dt +int_0^1 1/(4(1 - t)^2)dt + int_0^1 1/(2(1 + t))dt +int_0^1 1/(4(1 + t)^2)dt -int_0^1 1/(2(1 + t^2))dt`

`int_0^1 (t^2)/((1 - t^2)^2(1 + t^2)) dt = (1/2)ln(1-t)|_0^1 + (1/4)1/(1 - t)|_0^1 + (1/2)ln(1+t)|_0^1 + (1/4)1/(1+ t)|_0^1 + (1/2)tan^(-1) t|_0^1`

Using the fundamental theorem of calculus, yields:

`int_0^1 (t^2)/((1 - t^2)^2(1 + t^2)) dt = (1/2)ln(1-1) - (1/2)ln1 + (1/4)1/(1 - 1) - (1/4)1/1 + (1/2)ln(2) - (1/2)ln1 + 1/8 - 1/4 + pi/8 - 0`

You need to notice that ln 0 and 1/0 are invalid values, hence, the given integral cannot be evaluated under the given conditions.

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