Integration of sin2x/root a^2 - sin^4(x) , a>0

QUESTION:

\[ I = \int \frac{\sin 2x}{\sqrt{a^2 - \sin^4 x}} \, dx, \quad a > 0. \]

Step 1:We know that:

\[ \sin 2x = 2 \sin x \cos x. \]

Substituting this into the integral, we get:

\[ I = \int \frac{2 \sin x \cos x}{\sqrt{a^2 - \sin^4 x}} \, dx. \]

Step 2: Substitution \( \sin^2 x = t \)

Let \( \sin^2 x = t \). Then:

\[ 2 \sin x \cos x \, dx = dt, \quad \sin^4 x = t^2. \]

The integral now becomes:

\[ I = \int \frac{dt}{\sqrt{a^2 - t^2}}. \]

Step 3: Evaluate the integral

The integral \( \int \frac{dt}{\sqrt{a^2 - t^2}} \) is a standard result:

\[ I = \sin^{-1} \left( \frac{t}{a} \right) + C. \]

Step 4: Back-substitute \( t = \sin^2 x \)

Substituting \( t = \sin^2 x \) back, we get:

\[ I = \sin^{-1} \left( \frac{\sin^2 x}{a} \right) + C. \]

Final Answer:

\[ \boxed{I = \sin^{-1} \left( \frac{\sin^2 x}{a} \right) + C.} \]