integration of sqrt(sin(x))

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the indefinite integral `int sqrt(sin x) dx` , hence, you should come up with the following substitution, such that:

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

`sin x = (2t)/(1 + t^2)`

Changing the variable yields:

`int sqrt((2t)/(1 + t^2))*(dt)/(1 + t^2)`

You need to come up with the next substitution such that:

`1 + t^2 = u^2 => 2tdt = 2udu => dt = (2udu)/(2sqrt(u^2-1))`

`int sqrt((2sqrt(u^2-1)))*(du)/(u^2sqrt(u^2-1))` `= (sqrt2)int 1/(u^2(u^2-1)^(1/4)) du`

`u^2 - 1 = v^4> 2udu = 4v^3dv => du = (4v^3dv)/(2sqrt(v^4+1))`

`u^2 = v^4+1`

`(sqrt2)int 1/(v(v^4+1)) (4v^3dv)/(2sqrt(v^4+1))`

Since `(4v^3dv)/(2sqrt(v^4+1)) = (sqrt(v^4+1))'` , you may use the integration by parts, such that:

`(sqrt2)int 1/(v(v^4+1)) (sqrt(v^4+1))' dv`

`u = 1/(v^5+v) => du = -(5v^4+1)/((v^5+v)^2)`

`dv = (sqrt(v^4+1))' => v = sqrt(v^4+1)`

`int 1/(v(v^4+1)) (sqrt(v^4+1))' dv = (sqrt(v^4+1))/(v^5+v)+ int sqrt(v^4+1)*(5v^4+1)/((v^5+v)^2) dv`

`int sqrt(v^4+1)*(5v^4+1)/((v^5+v)^2) = int (5v^4+1)/(v(v^5+v)sqrt(v^4+1)) dv`

`(sqrt2)int 1/(v(v^4+1)) (sqrt(v^4+1))' dv =sqrt2(sqrt(v^4+1))/(v^5+v) + sqrt2 int (5v^4+1)/(v(v^5+v)sqrt(v^4+1)) dv`

Hence, integrating `sqrt sin x` , using the substitutions above and integration by parts yields  `int sqrt(sin x) dx = sqrt2(sqrt(v^4+1))/(v^5+v) + sqrt2 int (5v^4+1)/(v(v^5+v)sqrt(v^4+1)) dv.`

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