Evaluate ∫ (dx)/(sin⁴ x + cos⁴ x) — JEE Mathematics

Divide the numerator and denominator by $\cos^4 x$:

$$I = \int \frac{\sec^4 x}{\tan^4 x + 1} \, dx = \int \frac{\sec^2 x \cdot \sec^2 x}{\tan^4 x + 1} \, dx$$

Rewrite one $\sec^2 x$ as $1 + \tan^2 x$:

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

Let $t = \tan x$, then $dt = \sec^2 x \, dx$:

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

Divide the numerator and denominator by $t^2$:

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

Rewrite the denominator as $\left(t - \frac{1}{t}\right)^2 + 2$:

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

Let $u = t - \frac{1}{t}$, then $du = \left(1 + \frac{1}{t^2}\right) \, dt$:

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

Substitute back $u = t - \frac{1}{t} = \tan x - \cot x$:

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

Answer: $\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{\tan x - \cot x}{\sqrt{2}}\right) + C$