Find the indefinite integral of cos^4(x) with respect to x.

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TSR360
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#1
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#1
I can't seem to find a trig identity which simplifies the expression. I've tried the substitutions u = cos^2(x), u = cos(x), u = 1 - sin^2(x) & u = 1/2cos(2x) + 1/2 but du & u cannot replace all variables of x in the integral...
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mqb2766
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#2
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(Original post by TSR360)
I can't seem to find a trig identity which simplifies the expression. I've tried the substitutions u = cos^2(x), u = cos(x), u = 1 - sin^2(x) & u = 1/2cos(2x) + 1/2 but du & u cannot replace all variables of x in the integral...
You map (identity) to a multiple angle like
cos(4x) + ...
And integrate that.Just apply the double angle formula a couple of times.

Or try to get an identity where you can do the reverse chain rule (substitution).
Last edited by mqb2766; 1 year ago
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username2389033
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You can solve this problem by using a trigonometric identity.
Integral of cos^4 x dx
Integral of (cos^ 2 x)^2 dx
Since the based on trigonometric identity, the equivalent of cos^2 α= (1/2)(1+cos 2α) , let α=x and substitute this to integral equation. We have,
Integral of ((1/2)(1+cos2x))^2 dx
Integral of (1/4)(1 + 2 cos 2x +cos^2 2x dx)) dx
Then same process to the first note but the angle must be α=2x, and thus, cos^2 2x=(1/2)(1 + cos 2(2x))=(1/2)(1+cos 4x)
Then substitue the value of cos^2 2x to the preceeding solution.
Integral of 1/4(1+2cos 2x + (1/2)(1+cos4x)) dx.
Integrating these values would come up to:
3/8 x + 1/4 sin 2x + 1/32 sin 4x + C
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TSR360
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#4
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#4
(Original post by mqb2766)
You map (identity) to a multiple angle like
cos(4x) + ...
And integrate that.Just apply the double angle formula a couple of times.

Or try to get an identity where you can do the reverse chain rule (substitution).
Something like this: http://prntscr.com/w9l3r5 ?
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mqb2766
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(Original post by TSR360)
Something like this: http://prntscr.com/w9l3r5 ?
Yes (not fully checked), but it's a bit of a long way round.
cos^2(x) = (cos(2x)+1)/2
Square the right hand side and map cos^2(2x) to cos(4x) using the same identity.

Chuck the integral into wolframalpha for a quick check? I'm being lazy today :-)
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TeeEm
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(Original post by TSR360)
Something like this: http://prntscr.com/w9l3r5 ?
https://www.youtube.com/watch?v=uvZFXw9kjRY
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Eric Li
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#7
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Or maybe De Moivre's theorem can help.
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davros
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#8
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#8
(Original post by username2389033)
You can solve this problem by using a trigonometric identity.
Integral of cos^4 x dx
Please don't post full solutions - it's against the rules of the forum!

Have a good Christmas
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