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
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.
