The process of finding the antiderivative of a function is called integration.

## Answer: The solution of ∫ [(sin 2x) / (sin⁴ x + cos⁴ x)] dx = [ tanֿ¹ ( tan² x ) ] + C

Let's integrate [(sin 2x) / (sin⁴ x + cos⁴ x)] dx

**Explanation:**

We know that,

sin 2x = 2 sinx. cosx

So, ∫ [(sin 2x) / (sin⁴ x + cos⁴ x)] dx

= ∫ [(2 sinx. cosx) / (sin⁴ x + cos⁴ x)] dx

Now, dividing the numerator and denominator by cos⁴ x we get,

= ∫ [(2 sin x / cos³ x) / (tan⁴ x + 1)] dx

= ∫ [2 (sin x / cos x)(1 / cos² x) / (1 + tan⁴ x)] dx

= ∫ {1 / [1 + ( tan² x )²] } . 2 tan x sec² x dx

We get this in the form ∫ [1 / (1 + u²)] du

Where, u = tan² x, du = 2 tan x sec² x dx

From integration formula we have, ∫ [1 / (1 + u²)] du = [ tanֿ¹ (u) ] + C

Thus, ∫ {1 / [1 + ( tan² x )²] } . 2 tan x sec² x dx

= [ tanֿ¹ (tan² x) ] + C