What is `int_0^1 ` e^(arctgx)/(x^2+1)^(3/2)dx?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to re-write the integrand such that:

`int_0^1 e^(arctanx)/(x^2+1)^(3/2)dx = int_0^1 (e^(arctanx))/((x^2+1)sqrt(x^2 + 1)) dx`

You should notice that `(e^(arctanx))' = (e^(arctanx))/(x^2+1)` , hence, you may replace `(e^(arctanx))'` for `(e^(arctanx))/(x^2+1)` , such that:

`int_0^1 (e^(arctanx))/((x^2+1)sqrt(x^2 + 1)) dx = int_0^1 ((e^(arctanx))')/(sqrt(x^2 + 1)) dx`

You should use integration by parts, such that:

`int f(x)*g'(x)dx = f(x)*g(x) - int f'(x)*g(x)dx`

Let `1/(sqrt(x^2 + 1)) = f(x)` and `((e^(arctanx))') = g'(x)` , such that:

`f'(x) = -x/((x^2+1)sqrt(x^2 + 1))`

`g(x) = e^(arctanx)`

Using the formula yields:

`int_0^1 ((e^(arctanx))')/(sqrt(x^2 + 1)) dx = (e^(arctanx))/(sqrt(x^2 + 1))|_0^1 + int_0^1 (xe^(arctanx))/((x^2+1)sqrt(x^2 + 1)) dx`

`int_0^1 ((e^(arctanx))')/(sqrt(x^2 + 1)) dx = (e^(arctanx))/(sqrt(x^2 + 1))|_0^1 + int_0^1 ((e^(arctanx))')*x/(sqrt(x^2 + 1)) dx`

You need to solve the integral `int_0^1 ((e^(arctanx))')*x/(sqrt(x^2 + 1))dx` using parts again, such that:

`f(x) = x/(sqrt(x^2 + 1)) => f'(x) = (x^2 + 1 - x^2)/((x^2 + 1)sqrt(x^2 + 1))`

`f'(x) = 1/((x^2 + 1)sqrt(x^2 + 1))`

`g'(x) = ((e^(arctanx))') => g(x) = e^(arctanx)`

`int_0^1 ((e^(arctanx))')*x/(sqrt(x^2 + 1))dx = (xe^(arctanx))/(sqrt(x^2 + 1))|_0^1 - int_0^1 (e^(arctanx))/((x^2 + 1)sqrt(x^2 + 1)) dx`

Let `I = int_0^1 (e^(arctanx))/((x^2 + 1)sqrt(x^2 + 1)) dx` , hence, evaluating `I` , yields:

`I = (e^(arctanx))/(sqrt(x^2 + 1))|_0^1 + (xe^(arctanx))/(sqrt(x^2 + 1))|_0^1 - I`

Moving the terms that contain I to the left side, yields:

`2I = (e^(arctanx))/(sqrt(x^2 + 1))|_0^1 + (xe^(arctanx))/(sqrt(x^2 + 1))|_0^1`

By fundamental theorem of calculus, yields:

`2I = e^(arctan 1)/sqrt 2 - e^0/sqrt 1 + e^(arctan 1)/sqrt 2 - 0`

`2I = e^(pi/4)/sqrt2 - 1 + e^(pi/4)/sqrt2`

`I = e^(pi/4)/sqrt2 - 1/2`

Hence, evaluating the given definite integral, using integration by parts, yields `I = e^(pi/4)/sqrt2 - 1/2` .

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aruv | High School Teacher | (Level 2) Valedictorian

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`tan^(-1)x=t`

`x=tan(t)`

`1+x^2=1+tan^2(t)=sec^2(t)`

`sqrt(1+x^2)=sqrt(sec^2(t))=sec(t)=1/cos(t)`

`1/sqrt(1+x^2)=cos(t)`

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

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

`Thus`

`I=inte^(tan^(-1)(x))/(1+x^2)^(3/2)dx`

`=inte^(tan^(-1)(x))/((1+x^2)sqrt(1+x^2))dx`

`=int e^t cos(t) dt`

`=cos(t)e^t-inte^t (-sin(t))dt`

`=cos(t)e^t+sint e^t-inte^t cos(t)dt`

`=e^t(cos(t)+sin(t))-I`

`2I=e^t(cos(t)+sin(t))`

`I=(1/2)e^t (cos(t)+sin(t))`

`But`

`I=int_0^1e^(tan^(-1)(x))/(1+x^2)^(3/2)dx`

`tan^(-1)(1)=pi/4=t`

`tan^(-1)(0)=0=t`

Thus

`I=(1/2)e^t(cos(t)+sin(t))}_0^(pi/4)`

`=(1/2)(e^(pi/4)(1/sqrt(2)+1/sqrt(2))-e^0(1+0))`

`=(1/2)(sqrt(2)e^(pi/4)-1)`

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