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1018

  • one year ago

integrate cos^2(x) dx

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  1. UnkleRhaukus
    • one year ago
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    \[\int\cos^2x\,\mathrm dx = \int\frac{1+\cos2x}{2}\,\mathrm dx \\ \hspace{5em}= \frac12\int\,\mathrm dx +\frac12\int\cos2x\,\mathrm dx\\ \hspace{5em}=\]

  2. 1018
    • one year ago
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    hey thanks! but can you explain the first part? how did that become a fraction? is that a formula?

  3. 1018
    • one year ago
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    oh and also please the whole of it. haha. the 1 + and then if i should always bring down the exponent

  4. 1018
    • one year ago
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    thanks

  5. UnkleRhaukus
    • one year ago
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    from \[\cos(a)\cos(b)= \frac12\Big[\cos(a-b)+\cos(a+b)\Big]\] where, \(a=b=x\) \[\cos(x)\cos(x)= \frac12\Big[\cos(x-x)+\cos(x+x)\Big]\] \[\cos^2(x)= \frac12\Big[1+\cos(2x)\Big]\]

  6. UnkleRhaukus
    • one year ago
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    its probable a good idea to bring down the exponent with the 'Power Reducing Formula', (that i still have to look up every time) there might be another way

  7. UnkleRhaukus
    • one year ago
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    *probably

  8. 1018
    • one year ago
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    ok, i think i got it. i also looked up the power reducing formula. thanks!

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