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AI-তৈরি নমুনা উত্তর

We have:
$\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2 2x$
Using $\sin^2 2x = \frac{1 - \cos 4x}{2}$:
$\sin^4 x + \cos^4 x = 1 - \frac{1 - \cos 4x}{4} = \frac{3 + \cos 4x}{4}$

Thus, the integral becomes:
$I = \int \frac{\sin 4x}{\sin^4 x + \cos^4 x} dx = \int \frac{4\sin 4x}{3 + \cos 4x} dx$

Let $u = 3 + \cos 4x$.
Then $du = -4\sin 4x dx \implies 4\sin 4x dx = -du$.

Substituting these into the integral:
$I = \int \frac{-du}{u} = -\ln|u| + C = -\ln(3 + \cos 4x) + C$
(where $C$ is the constant of integration).