Calculate integral (x² dy)/(x^(5/3)+y^(5/3)) and integral

Calculate $\int \frac{x^2 \, dy}{x^{5/3}+y^{5/3}}$ and $\int \frac{y^2 \, dx}{x^{5/3}+y^{5/3}}$ if $x = a\cos^3 t$, $y = a\sin^3 t$; $t \in [0, \frac{\pi}{2}]$.

We will use the given parametrization to evaluate both integrals.

Given parametrization

\[x = a\cos^3 t, \quad y = a\sin^3 t, \quad t \in \left[0, \frac{\pi}{2}\right]\]

This is the parametrization of an astroid (or hypocycloid with four cusps).

Step 1: Find $dx$ and $dy$

\[dx = \frac{dx}{dt} \, dt = 3a\cos^2 t \cdot (-\sin t) \, dt = -3a\cos^2 t \sin t \, dt\]

\[dy = \frac{dy}{dt} \, dt = 3a\sin^2 t \cdot \cos t \, dt = 3a\sin^2 t \cos t \, dt\]

Step 2: Compute $x^{5/3}$ and $y^{5/3}$

\[x^{5/3} = (a\cos^3 t)^{5/3} = a^{5/3} \cos^5 t\]

\[y^{5/3} = (a\sin^3 t)^{5/3} = a^{5/3} \sin^5 t\]

\[x^{5/3} + y^{5/3} = a^{5/3}(\cos^5 t + \sin^5 t)\]

Part 1: Calculate $\int \frac{x^2 \, dy}{x^{5/3}+y^{5/3}}$

Step 3: Substitute into the integral

\[\int \frac{x^2 \, dy}{x^{5/3}+y^{5/3}} = \int_0^{\pi/2} \frac{(a\cos^3 t)^2 \cdot 3a\sin^2 t \cos t \, dt}{a^{5/3}(\cos^5 t + \sin^5 t)}\]

\[= \int_0^{\pi/2} \frac{a^2 \cos^6 t \cdot 3a\sin^2 t \cos t}{a^{5/3}(\cos^5 t + \sin^5 t)} \, dt\]

\[= \int_0^{\pi/2} \frac{3a^3 \cos^7 t \sin^2 t}{a^{5/3}(\cos^5 t + \sin^5 t)} \, dt\]

\[= 3a^{3-5/3} \int_0^{\pi/2} \frac{\cos^7 t \sin^2 t}{\cos^5 t + \sin^5 t} \, dt\]

\[= 3a^{4/3} \int_0^{\pi/2} \frac{\cos^7 t \sin^2 t}{\cos^5 t + \sin^5 t} \, dt\]

Step 4: Factor the numerator

\[\frac{\cos^7 t \sin^2 t}{\cos^5 t + \sin^5 t} = \frac{\cos^5 t \cdot \cos^2 t \sin^2 t}{\cos^5 t + \sin^5 t}\]

Let's try a different approach. Notice that:

\[\frac{\cos^7 t \sin^2 t}{\cos^5 t + \sin^5 t} = \frac{\cos^5 t \cdot \cos^2 t \sin^2 t}{\cos^5 t + \sin^5 t}\]

Actually, let me use symmetry. Let $I_1 = \int_0^{\pi/2} \frac{\cos^7 t \sin^2 t}{\cos^5 t + \sin^5 t} \, dt$.

Part 2: Calculate $\int \frac{y^2 \, dx}{x^{5/3}+y^{5/3}}$

\[\int \frac{y^2 \, dx}{x^{5/3}+y^{5/3}} = \int_0^{\pi/2} \frac{(a\sin^3 t)^2 \cdot (-3a\cos^2 t \sin t)}{a^{5/3}(\cos^5 t + \sin^5 t)} \, dt\]

\[= -\int_0^{\pi/2} \frac{a^2 \sin^6 t \cdot 3a\cos^2 t \sin t}{a^{5/3}(\cos^5 t + \sin^5 t)} \, dt\]

\[= -3a^{4/3} \int_0^{\pi/2} \frac{\sin^7 t \cos^2 t}{\cos^5 t + \sin^5 t} \, dt\]

Let $I_2 = \int_0^{\pi/2} \frac{\sin^7 t \cos^2 t}{\cos^5 t + \sin^5 t} \, dt$.

Step 5: Use symmetry

Using the substitution $u = \frac{\pi}{2} - t$ (which swaps $\sin$ and $\cos$):

\[I_2 = \int_0^{\pi/2} \frac{\cos^7 u \sin^2 u}{\sin^5 u + \cos^5 u} \, du = I_1\]

Therefore: $I_1 = I_2$

Step 6: Add the integrals

\[I_1 + I_2 = \int_0^{\pi/2} \frac{\cos^7 t \sin^2 t + \sin^7 t \cos^2 t}{\cos^5 t + \sin^5 t} \, dt\]

\[= \int_0^{\pi/2} \frac{\cos^2 t \sin^2 t(\cos^5 t + \sin^5 t)}{\cos^5 t + \sin^5 t} \, dt\]

\[= \int_0^{\pi/2} \cos^2 t \sin^2 t \, dt\]

\[= \int_0^{\pi/2} \frac{\sin^2(2t)}{4} \, dt = \frac{1}{4} \int_0^{\pi/2} \sin^2(2t) \, dt\]

Using $\sin^2\theta = \frac{1-\cos(2\theta)}{2}$:

\[= \frac{1}{4} \int_0^{\pi/2} \frac{1-\cos(4t)}{2} \, dt = \frac{1}{8}\left[t - \frac{\sin(4t)}{4}\right]_0^{\pi/2}\]

\[= \frac{1}{8}\left[\frac{\pi}{2} - 0\right] = \frac{\pi}{16}\]

Since $I_1 = I_2$: $2I_1 = \frac{\pi}{16}$, so $I_1 = I_2 = \frac{\pi}{32}$

Final answers:

\[\boxed{\int \frac{x^2 \, dy}{x^{5/3}+y^{5/3}} = 3a^{4/3} \cdot \frac{\pi}{32} = \frac{3\pi a^{4/3}}{32}}\]

\[\boxed{\int \frac{y^2 \, dx}{x^{5/3}+y^{5/3}} = -3a^{4/3} \cdot \frac{\pi}{32} = -\frac{3\pi a^{4/3}}{32}}\]

Note: The negative sign in the second integral comes from the $dx = -3a\cos^2 t \sin t \, dt$ term.