Question:
The value of \(\displaystyle\int^{\frac{\pi}{16}}_{0}\cos6x\cos2xdx\) is equal to
The value of \(\displaystyle\int^{\frac{\pi}{16}}_{0}\cos6x\cos2xdx\) is equal to
Updated On: May 31, 2024
- \(\frac{1+\sqrt 2}{16}\)
- \(\frac{1+\sqrt 2}{8}\)
- \(\frac{2+\sqrt 2}{16}\)
- \(\frac{-1+\sqrt 2}{16}\)
- \(\frac{-1+\sqrt 2}{8}\)
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The Correct Option is A
Solution and Explanation
The correct option is (A) : \(\frac{1+\sqrt 2}{16}\)
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