Polar coordinates question (integration) Coming to the end of a question on Polar Coordinates and I have come across this integral which I cannot seem to evaluate. Upon expanding the entire expression, I'm not sure how to use the formula $cos^2(\theta) = \frac{1 +cos(2 \theta)}{2}$ in order to integrate the powers without everything becoming absurdly messy. Happy to accept hints.
Book is Calculus of Several Variables by Serge Lang

 A: Note that the odd powers have integral zero (proof at the end). You have
\begin{align}
\int_0^{2\pi}(1-\cos t)^4\cos^2t\,dt
&=\int_0^{2\pi}(\cos^6t-4\cos^5t+6\cos^4t-4\cos^3t+\cos^2t)\,dt\\[0.3cm]
&=\int_0^{2\pi}(\cos^6t+6\cos^4t+\cos^2t)\,dt\\[0.3cm]
\end{align}
Now, without calculating a single antiderivative,
$$
\int_0^{2\pi}\cos^2t\,dt=\frac12\,\int_0^{2\pi}(1+\cos2t)\,dt=\frac12\int_0^{2\pi}1\,dt=\pi.
$$
And
\begin{align}
\int_0^{2\pi}\cos^4t\,dt&=\frac14\,\int_0^{2\pi}(1+\cos2t)^2\,dt \\[0.3cm]
&=\frac14\,\int_0^{2\pi}(1+2\cos2t+\cos^22t)\,dt \\[0.3cm]
&=\frac14\,\int_0^{2\pi}(1+\cos^22t)\,dt \\[0.3cm]
&=\frac14\,(2\pi+\pi)=\frac{3\pi}4.
\end{align}
And
\begin{align}
\int_0^{2\pi}\cos^6t\,dt&=\frac18\,\int_0^{2\pi}(1+\cos2t)^3\,dt \\[0.3cm]
&=\frac18\,\int_0^{2\pi}(1+3\cos2t+3\cos^22t+\cos^32t)\,dt \\[0.3cm]
&=\frac18\,\int_0^{2\pi}(1+3\cos^2t)\,dt \\[0.3cm]
&=\frac18\,(2\pi+3\pi)=\frac{5\pi}8.
\end{align}
Adding,
$$
\int_0^{2\pi}(1-\cos t)^4\cos^2t\,dt
=\pi+\frac{6\times 3\pi}4+\frac{5\pi}8=\frac{49}8.
$$

Proof that the odd powers have integral zero.
\begin{align}
\int_0^{2\pi}\cos^{2k+1}t \,dt
&=\int_{-\pi}^{\pi}\cos^{2k+1}t \,dt
=\int_{-\pi}^{\pi}\sin^{2k+1}(t+\tfrac\pi2)\,dt\\[0.3cm]
&=\int_{-\pi+\tfrac\pi2}^{\pi+\tfrac\pi2}\sin^{2k+1}t\,dt\\[0.3cm]
&=\int_{-\pi+\tfrac\pi2}^{\pi}\sin^{2k+1}t\,dt
+\int_{\pi}^{\pi+\tfrac\pi2}\sin^{2k+1}t\,dt\\[0.3cm]
&=\int_{-\pi+\tfrac\pi2}^{\pi}\sin^{2k+1}t\,dt
+\int_{-\pi}^{-\pi+\tfrac\pi2}\sin^{2k+1}t\,dt\\[0.3cm]
&=\int_{-\pi}^{\pi}\sin^{2k+1}t\,dt=0,
\end{align}
since an odd power of the sine is an odd function.
