Evaluate the following definite integrals: _ 0 ^ 6 2x dx - Vidyadip

Question

Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{6}}\cos\text{x }\cos2\text{x}\text{ dx}$

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Answer

We have,
$\int_{0}^\limits{\frac{\pi}{6}}\cos\text{x }\cos2\text{x}\text{ dx}$ $\big[\because2\cos\text{C}\cos\text{D}=\cos(\text{C}+\text{D})-\cos(\text{C}-\text{D})\big]$
$=\frac{1}{2}\int_{0}^\limits{\frac{\pi}{6}}2\cos\text{x }\cos2\text{x dx}$
$=\frac{1}{2}\int_{0}^\limits{\frac{\pi}{6}}(\cos3\text{x}+\cos\text{x})\text{dx}$
$=\frac{1}{2}\int\Big[\frac{\sin3\text{x}}{3}+\sin\text{x}\Big]_0^{\frac{\pi}{6}}$
$=\frac{1}{2}\Bigg[\bigg(\frac{\sin3\frac{\pi}{6}}{3}+\sin\frac{\pi}{6}\bigg)-(\sin0-\sin0)\Bigg]$
$=\frac{1}{2}\bigg[\frac{\sin\frac{\pi}{2}}{3}+\sin\frac{\pi}{6}\bigg]$
$=\frac{1}{2}\Big(\frac{1}{3}+\frac{1}{2}\Big)$
$=\frac{1}{2}\Big(\frac{5}{6}\Big)$
$=\frac{5}{12}$
$\therefore\ \int_{0}^\limits{\frac{\pi}{6}}\cos\text{x }\cos2\text{x}\text{ dx}=\frac{5}{12}$

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