If dx/^4 x + ^4 x = {1}{{2}} ^{-1}[g(x)] + c, then g(x) equals _______

Correct Answer is : tan⁡x−cot⁡x2{ x - x}{{2}}2tanx−cotx

Correct Answer is : tan⁡x−cot⁡x2{ x - x}{{2}}2tanx−cotx

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AP EAPCET 23-Aug-2021 Shift 1 Paper

Section: Mathematics

Question : 77 of 160

Marks: +1, -0

\int \frac{dx}{\cos^4 x + \sin^4 x} = \frac{1}{\sqrt{2}} \tan^{-1}[g(x)] + c

g(x)

equals _______

Solution:

I = \int \frac{dx}{\cos^4 x + \sin^4 x}

= \int \frac{\sec^4 x}{1 + \tan^4 x} dx

= \int \frac{\sec^2 x (1+\tan^2 x)}{1+\tan^4 x} dx

Let

\tan x = t

, then

\sec^2 x \, dx = dt

I = \int \frac{1+t^2}{1+t^4} dt

= \int \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} dt

= \int \frac{1 + \frac{1}{t^2}}{\left(t - \frac{1}{t}\right)^2 + 2} dt

Let

t - \frac{1}{t} = u

, then

\left(1 + \frac{1}{t^2}\right) dt = du

I = \int \frac{du}{u^2+2} = \int \frac{du}{u^2 + (\sqrt{2})^2}

= \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{u}{\sqrt{2}}\right) + C

I = \frac{1}{\sqrt{2}} \tan^{-1}\left( \frac{t - \frac{1}{t}}{\sqrt{2}} \right) + C

= \frac{1}{\sqrt{2}} \tan^{-1}\left( \frac{t^2-1}{\sqrt{2} \, t} \right) + C

I = \frac{1}{\sqrt{2}} \tan^{-1}\left( \frac{\tan^2 x - 1}{\sqrt{2} \tan x} \right) + C

\therefore g(x) = \frac{\tan^2 x - 1}{\sqrt{2} \tan x}

= \frac{1}{\sqrt{2}} (\tan x - \cot x)

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