If dx cos^4 ⁡x+ sin^4 ⁡x= 1 2 tan^-1 ⁡ [g (x ) ]+C, then g (x ) equals

Question

Accepted Answer

Correct answer: A. (a) Let , dx cos^4 ⁡x+ sin^4 ⁡x=II= sec^4 ⁡x1+ tan^4 ⁡x dx= sec^2 ⁡x(1+ tan^2 ⁡x)1+ tan^4 ⁡x Let tan ⁡x=t, then sec^2 ⁡ xdx= dt 1= 1+t^21+t^4 dt= (1+ 1t^2)(t^2+ 1t^2)dt Let t- 1t=u, then (1+ 1t^2) dt= du Now, I= duu^2+2= 1 2 tan^-1 ⁡ ( u 2 )+C I= 1 2 tan^-1 ⁡ ( t- 1t 2 )+CI= 1 2 tan^-1 ⁡ ( tan^2 ⁡x-1 2 tan ⁡x )+C Comparing it with 1 2 tan^-1 ⁡ (g (x ) )+c we get ∴g (x )= tan^2-1 2 tan ⁡x= 1 2 ( tan ⁡x- cot ⁡x )

$If \int \frac{dx}{cos ^{4} x + sin ^{4} x} = \frac{1}{2} tan^{- 1} [g (x)] + C, then g (x) equals$

Solution

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