<s>[INST] Evaluate the integral with and without Green's theorem Evaluate the line integral $\oint_C y^2dx + xdy$ when $C$ has the vector equation $\alpha(t)=(2\cos^3t)i+(2\sin^3t)j$, $0 \leq t \leq 2\pi$.
My attempt
BY USING GREEN's THEOREM,
i.e., $P=y^2$, $Q=x$
we get $\iint_c 1-2y dxdy$ 
Now putting $x=r\cos^3t$ and $y=r\sin^3t$ 
we get $dxdy=Jdrdt$, where $J=3r \sin^2t \cos^2t$
hence $dxdy=3r \sin^2t \cos^2t drdt$
Hence the required integral is,
$\int_0^{2\pi} \int_0^2 (1-2r \sin^3t)(3r \sin^2t \cos^2t)drdt$
which is equal to $\pi /4$
But the answer given in APOSTOL's is $3 \pi /2$
NOW BY NORMAL METHOD 
put $x=2 \cos^3t$ and $y=2 \sin^3t$
$dx=-6 \cos^2t \sin t dt$ and $dy=6 \sin^2t cost dt$,
Hence the integral becomes $-24 \int_0^{2 \pi} \sin^7t \cos^2t dt + 12 \int_0^{2 \pi} \cos^4t \sin^2t dt$ which is equal to ZERO.
WHAT MISTAKE AM I DOING AND THE CORRECT ANSWER IS $3 \pi /2$  ?
 [/INST] Your second method
Use $$\int_0^{2\pi}\sin^n x\ dx=\dfrac{2\pi}{2^{2k}}{2k\choose k}$$
for even $n=2k$ and for odd $n$ it is zero, then
$$-24 \int_0^{2 \pi} \sin^7t \cos^2t dt + 12 \int_0^{2 \pi} \cos^4t \sin^2t dt$$
$$-24\int_0^{2 \pi} \sin^7t (1-\sin^2t) dt + 12 \int_0^{2 \pi} (1-\sin^2t)^2\sin^2t dt$$
$$-24\int_0^{2 \pi} \sin^7t\ dt + 24\int_0^{2 \pi} \sin^9t\ dt + 12\int_0^{2 \pi} \sin^2t\ dt -24 \int_0^{2 \pi} \sin^4t\ dt + 12\int_0^{2 \pi} \sin^6t\ dt $$
$$0+0+12\dfrac{2\pi}{2^{2}}{2\choose 1} -24\dfrac{2\pi}{2^{4}}{4\choose 2} + 12\dfrac{2\pi}{2^{6}}{6\choose 3}=\color{blue}{\dfrac{3\pi}{2}}$$
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