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प्रश्न
Evaluate : `∫1/(cos^4x+sin^4x)dx`
उत्तर
Let I = `int 1/(cos^4x+sin^4x) dx`
Divide numerator and denominator by cos4x, we get:
`int [sec^4x]/[1 + tan^4x]` dx
`int [sec^2 x(sec^2x)]/[ 1 + tan^4x ]` dx
`int [sec^2 x( 1 + tan^2 x)]/( 1 + tan^4x )`dx
Putting tan x = t,
Sec2x dx = dt
I = `int ( 1 + t^2)/(1+ t^4) dt`
Dividing the numerator and denominator by t2, we get:
I = `int [ 1 + t^(1/2) ]/[ t^(1/2) + t^2 ]`
I = `int [ 1 + 1/t^2]/[(t - 1/t)^2 + 2]` dt
Let t - `1/t` = u
`1 + 1/t^2 = (du)/dt`
`( 1 + 1/t^2) dt = du`
I = `1/sqrt2 tan^-1 (u/sqrt2) + C`
I = `1/sqrt2 tan^-1 (( t - 1/t )/sqrt2) + C`
I = `1/sqrt2 tan^-1 ((t^2 - 1)/(sqrt2t)) + C`
I = `1/sqrt2 tan^-1 ( tan^2x - 1)/(sqrt2tan x ) + C`.
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