Question:

Step 1: Set up the integral and simplify by dividing numerator and denominator by $\cos^8 x$. We start with: $$I = \int_0^{\pi/4} \frac{\sin^3 x \cos^3 x}{(\sin^4 x + \cos^4 x)^2}\,dx$$ Dividing both numerator and denominator by $\cos^8 x$: $$I = \int_0^{\pi/4} \frac{\frac{\sin^3 x \cos^3 x}{\cos^8 x}}{\frac{\sin^4 x + \cos^4 x}{\cos^8 x}}\,dx = \int_0^{\pi/4} \frac{\tan^3 x \sec^2 x}{(\tan^4 x + 1)^2}\,dx$$ Step 2: Apply the substitution $t = \tan^2 x$. Let $t = \tan^2 x$, which gives us $dt = 2\tan x \sec^2 x\,dx$. Determining the limits: - When $x = 0$: $t = \tan^2(0) = 0$ - When $x = \pi/4$: $t = \tan^2(\pi/4) = 1$ We can rewrite $\tan^3 x \sec^2 x\,dx$ as: $$\tan^3 x \sec^2 x\,dx = \tan^2 x \cdot \tan x \sec^2 x\,dx = t \cdot \frac{dt}{2}$$ Step 3: Transform the integral using the substitution. Substituting into the integral: $$I = \int_0^1 \frac{t}{(t^2+1)^2} \cdot \frac{dt}{2} = \frac{1}{2}\int_0^1 \frac{t\,dt}{(t^2+1)^2}$$ Step 4: Apply a second substitution $u = t^2 + 1$. Let $u = t^2 + 1$, which gives us $du = 2t\,dt$. Determining the new limits: - When $t = 0$: $u = 1$ - When $t = 1$: $u = 2$ The integral becomes: $$I = \frac{1}{2} \cdot \int_1^2 \frac{1}{u^2} \cdot \frac{du}{2} = \frac{1}{4}\int_1^2 \frac{du}{u^2}$$ Step 5: Evaluate the integral. $$I = \frac{1}{4}\int_1^2 u^{-2}\,du = \frac{1}{4}\left[-\frac{1}{u}\right]_1^2$$ $$I = \frac{1}{4}\left(-\frac{1}{2} - \left(-\frac{1}{1}\right)\right) = \frac{1}{4}\left(-\frac{1}{2} + 1\right) = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}$$ **Final Answer:** The value of the definite integral is $\boxed{\dfrac{1}{8}}$, which corresponds to **Option 4**.