how calculate integral sign(lower lim=0,upper limit=pi/4)1/(sin^2x+4cos^2x+2)dx? how change x? cos x=t or sin x=t? 

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to factor out `cos^2 x`  to denominator such that:

`1/(sin^2 x+ 4cos^2 x + 2) = 1/((cos^2 x)((sin^2 x)/(cos^2 x) + 4 + 2/(cos^2 x)))`

You should remember that you may substitute `1 + tan^2 x`  for `1/(cos^2 x)`  such that:

`1/((cos^2 x)(tan^2 x+ 4 + 2(1 + tan^2 x))) = (1/(cos^2 x))/(3tan^2 x + 6)`

Hence, evaluating the definite integral yields:

`int_0^(pi/4) 1/(sin^2 x + 4cos^2 x + 2) dx = int_0^(pi/4) (1/(cos^2 x))/(3tan^2 x + 6)`

You need to remember that `(tan x)' = 1/(cos^2 x)`

You need to substitute t for tan x such that:

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

You need to remember that if you change the variable, you also need to change the limits of integration such that:

`x = 0 => tan 0 = 0`

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

`int_0^1 (dt)/(3t^2 + 6) = (1/3)int_0^1 (dt)/(t^2 + 2)`

`int_0^1 (dt)/(3t^2 + 6) = (1/3)*(1/sqrt2)*(tan^(-1)t/sqrt2)|_0^1`

`int_0^1 (dt)/(3t^2 + 6) = 1/(3sqrt2)(tan^(-1) sqrt2/2 - tan^(-1) 0)`

`int_0^1 (dt)/(3t^2 + 6) = 1/(3sqrt2)(tan^(-1) sqrt2/2)`

Hence, evaluating the given definite integral, using the substitution `tan x = t` , yields `int_0^(pi/4) 1/(sin^2 x + 4cos^2 x + 2) dx = 1/(3sqrt2)(tan^(-1) sqrt2/2).`

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