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Evaluate: $\int\limits_{0}^{\large\frac{\pi}{4}}\cos^{8}2x dx$

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  • $\int \limits_0^{\large\frac{\pi}{2}} \sin ^n x dx$$=\int \limits_0^{\large\frac{\pi}{2}} \cos ^n x dx= \left\{ \begin{array}{1 1} \large\frac{n-1}{n}.\frac{n-3}{n-2}.\frac{n-5}{n-4}...\frac{2}{3}\; \normalsize when\; n\; is\; odd \\ \large\frac{n-1}{n}.\frac{n-3}{n-2}.\frac{n-5}{n-4}...\frac{1}{2}\frac{\pi}{2} \normalsize \;when\; n\; is\; even\; \end{array} \right. $
$\int\limits_{0}^{\large\frac{\pi}{4}}\cos^{8}2x dx$
Step 1:
Let $ 2x=t$
$2dx=dt=>dx=\large\frac{dt}{2}$
When $x=0\qquad t=0$
When $x=\large\frac{\pi}{6},\qquad t=\large\frac{\pi}{2}$
Step 2:
$\int\limits_{0}^{\large\frac{\pi}{4}}\cos 2x dx=\large\frac{1}{2} \int \limits _0^{\large\frac{\pi}{2}}$$ \cos ^8 t dt $
$\qquad=\large\frac{1}{2}.\frac{7}{8}.\frac{5}{6}.\frac{3}{4}.\frac{1}{2}.\frac{\pi}{2}$
$\qquad=\large\frac{35 \pi}{512}$
answered Aug 14, 2013 by meena.p
 

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