1 problem found
Prove the identities \(\cos^4\theta -\sin^4\theta \equiv \cos 2\theta\) and $\cos^4 \theta + \sin^4 \theta \equiv 1 - {\frac12} \sin^2 2 \theta$. Hence or otherwise evaluate \[ \int_0^{\frac{1}{2}\pi} \cos^4 \theta \; \d \theta \;\;\;\; \mbox{and}\;\;\;\; \int_0^{\frac{1}{2}\pi} \sin^4 \theta \; \d \theta \,. \] Evaluate also \[ \int_0^{\frac{1}{2}\pi} \cos^6 \theta \; \d \theta \;\;\;\; \mbox{and}\;\;\;\; \int_0^{\frac{1}{2}\pi} \sin^6 \theta \; \d \theta \,. \]
Solution: \begin{align*} && \cos^4 \theta - \sin^4 \theta &= (\cos^2 \theta - \sin^2 \theta)(\cos^2 \theta + \sin^2 \theta) \\ &&&= \cos^2 \theta - \sin^2 \theta \\ &&&= \cos 2 \theta \\ \\ && 1&= (\cos^2 \theta + \sin^2 \theta)^2 \\ &&&= \cos^4 \theta + \sin^4 \theta + 2 \sin^2 \theta \cos^2 \theta \\ &&&= \cos^4 \theta + \sin^4 \theta + \frac12 ( \sin^2 2 \theta) \\ \Rightarrow && \cos^4 \theta + \sin^4 \theta &= 1 - \tfrac12 \sin^2 2 \theta \end{align*} \begin{align*} && I &= \int_{0}^{\pi/2} \cos^4 \theta \d \theta \\ && J &= \int_0^{\pi/2} \sin^4 \theta \d \theta \\ && I-J &= \int_0^{\pi/2} \cos 2 \theta \d \theta = 0 \\ && I+J &= \int_0^{\pi/2} (1- \frac12 \sin^2 2 \theta) \d \theta \\ &&&= \frac{\pi}{2} - \frac14 \int_0^{\pi} \sin^2 \theta \d \theta \\ &&&= \frac{\pi}{2} - \frac{\pi}{8} \\ &&&= \frac{3\pi}{8} \\ \Rightarrow && I=J &= \frac{3\pi}{16} \end{align*} \begin{align*} && \cos^6 \theta + \sin^6 \theta &= (\cos^2 \theta + \sin^2 \theta)(\cos^4 \theta - \cos^2 \theta \sin^2 \theta + \sin^4 \theta) \\ &&&= 1-\tfrac12 \sin^2 2\theta - \tfrac14 \sin^2 2 \theta \\ &&&= 1 - \tfrac34 \sin^2 2 \theta \\ %&& \cos^6 \theta - \sin^6 \theta &= (\cos^2 \theta - \sin^2 \theta)(\cos^4 \theta + \cos^2 \theta \sin^2 \theta + \sin^4 \theta) \\ %&&&= \cos 2 \theta (1 - \tfrac12 \sin^2 2 \theta + \tfrac14 \sin^2 2 \theta) \\ %&&&= \cos 2 \theta (1 - \tfrac14 \sin^2 2 \theta) \\ \end{align*} \begin{align*} && I &= \int_{0}^{\pi/2} \cos^6 \theta \d \theta \\ && J &= \int_0^{\pi/2} \sin^6 \theta \d \theta \\ && I-J &= 0 \\ && I+J &= \int_0^{\pi/2} (1 - \tfrac34 \sin^2 2 \theta) \d \theta \\ &&&= \frac{\pi}{2} - \frac{3\pi}{16} = \frac{5\pi}{16} \\ \Rightarrow && I = J &= \frac{5\pi}{32} \end{align*}