Is there a hand-written way of integrating sec^2(x)cos^8(x) USING calculus one methods? No product rule.

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Is there a hand-written way of integrating sec^2(x)cos^8(x) USING calculus one methods? No product rule.

Mathematics
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Recall that \[sec\ x = \frac{1}{cos\ x}\] So you can just cancel 2 of the cosines.
Yes but they're to different powers.
Yes but \[x^8 * \frac{1}{x^2} = x^6\]

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I didn't think you could cancel say cos(5x2)*1/cos(x^8) and get their inner functions to come outside.
You cannot. But \[cos^8x \ne cos(x^8)\]
\[cos^8x = (cos\ x)^8\]
oh i see, so after cancel how would I integrate cos^6(x)?
Use the fact that \[cos^2(x) = \frac{1 + cos(2x)}{2}\] So \[cos^6x = (cos^2x)^3 = [\frac{1 + cos(2x)}{2}]^3 = \frac{1 + cos(2x)}{2}*\frac{1 + cos(2x)}{2}*\frac{1 + cos(2x)}{2}\] \[= \frac{1}{8}(1+ 2cos(2x) + cos^2(2x))(1+cos(2x))\]
And then distribute again, then use the rule again to break down the other even powers of cos(2x) into cos(4x) etc.
It quickly becomes a mess.
But it's doable.
God that sounds terrible. I took a calc test earlier today and had two problems on it that were like that. The other was one like cos(x^3)sin(x) and was an indefinite integral. I put it was unsolvable :P
Was it \(cos(x^3)sin(x)\) or was it \((cos^3x)sin(x)\)
i believe it was the first one. which can't be u-subbed, so I had no idea what to do.
Right, but the other one can be u-subbed.
So have to be sure you know which they're talking about.
I'm quite bad at using algebra on trig equations. I understand trig but algebraically transforming them is unknown to me.

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