Hello, I would like to know if you can help me with this problem the intergal of sqrt(cos (6x)+1), I’ve understood it as far as cos (1/2 (ax)) but where is the sqrt of 2 coming from?! Thank you.
Find the indefinite integral:
. . . . .\(\displaystyle \displaystyle \int\, \bigg(\sqrt{\strut \cos(6x)\, +\, 1\,}\bigg)\, dx\)
Step (1)
Apply rule:
. . . . .\(\displaystyle \displaystyle \int\, \bigg(\sqrt{\strut \cos(ax)\, +\, 1\,}\bigg)\, dx\, \longrightarrow\, \int\, \bigg(\sqrt{\strut 2\,}\, \cos\left(\dfrac{1}{2}ax\right)\bigg)\, dx\)
. . . . .\(\displaystyle \displaystyle \int\, \bigg(\sqrt{\strut \cos(6x)\, +\, 1\,}\bigg)\, dx\, =\, \dfrac{1}{3}\, \sqrt{\strut 2\,}\, \sin(3x)\)
Find the indefinite integral:
. . . . .\(\displaystyle \displaystyle \int\, \bigg(\sqrt{\strut \cos(6x)\, +\, 1\,}\bigg)\, dx\)
Step (1)
Apply rule:
. . . . .\(\displaystyle \displaystyle \int\, \bigg(\sqrt{\strut \cos(ax)\, +\, 1\,}\bigg)\, dx\, \longrightarrow\, \int\, \bigg(\sqrt{\strut 2\,}\, \cos\left(\dfrac{1}{2}ax\right)\bigg)\, dx\)
. . . . .\(\displaystyle \displaystyle \int\, \bigg(\sqrt{\strut \cos(6x)\, +\, 1\,}\bigg)\, dx\, =\, \dfrac{1}{3}\, \sqrt{\strut 2\,}\, \sin(3x)\)
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