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My calculus skills are a bit rusty and I am working trig integrals. I checked here, Help with $\int \cos^6 x dx$ but the way I go about solving this is a bit different. So computing $\int \cos^6x \, dx$ what I do is, $$\int \cos^6 dx \, = \, \int \cos^4 x \cos^2 x \, dx$$ $$\int \cos^4 x \cos^2 x \, dx \, = \, \int \left(\frac {1+\cos 2x}2\right)^2\left(\frac {1+\cos 2x}2\right)dx$$ $$=\, \frac 18 \int \left(1+3\cos 2x + 3\cos^2 2x+ \cos^3 2x\right)dx$$ I split this into 3 integrals and simplify them, $$\frac 18 \int 1+3\cos 2x \, dx\;+\; \frac 3{8}\int \cos^2 2x\,dx \;+\; \frac 1{8} \int \cos^3 2x\, dx$$ $$= \, \frac 18 \int 1+3\cos 2x \, dx\;+\; \frac 3{16}\int 1+\cos 4x\,dx \;+\; \frac 1{8}\int \cos^2 2x \cos 2x \, dx$$ $$= \, \frac 18 \int 1+3\cos 2x \, dx\;+\; \frac 3{16}\int 1+\cos 4x\,dx \;+\; \frac 1{16}\int \left(1+\cos 4x\right) \cos 2x \, dx$$

Edit (correct answer found)

$$= \, \frac 18 \int 1+3\cos 2x \, dx\;+\; \frac 3{16}\int 1+\cos 4x\,dx \;+\; \frac 1{16}\int \cos 2x dx + \frac 1{16} \int \cos 4x \cos 2x \, dx$$ $$= \, \frac 18 \int 1+3\cos 2x \, dx\;+\; \frac 3{16}\int 1+\cos 4x\,dx \;+ \frac 1{16}\int \cos 2x dx +\; \frac 1{32}\int \cos 2x + \cos 6x \, dx$$ Solving one at a time I get, $$\frac x8 + \frac 3{16}\sin 2x \;+\; \frac 3{16}\int 1+\cos 4x\,dx \;+ \frac 1{16}\int \cos 2x dx +\; \frac 1{32}\int \cos 2x + \cos 6x \, dx$$ $$= \, \frac x8 + \frac 3{16}\sin 2x \;+\; \frac 3{16}x + \frac 3{64} \sin 4x \;+\frac 1{16}\int \cos 2x dx +\; \frac 1{32}\int \cos 2x + \cos 6x \, dx$$ $$= \, \frac x8 + \frac 3{16}\sin 2x \;+\; \frac 3{16}x + \frac 3{64} \sin 4x \;+\frac 1{32}\sin 2x dx +\; \frac 1{32}\int \cos 2x + \cos 6x \, dx$$ $$= \, \frac x8 + \frac 3{16}\sin 2x \;+\; \frac 3{16}x + \frac 3{64} \sin 4x \;+\frac 1{32}\sin 2x \, +\; \frac 1{64} \sin 2x + \frac 1{192} \sin 6x$$ Finally, $$= \, \frac 5{16}x + \frac {15}{64}\sin 2x + \frac 3{64} \sin 4x + \frac 1{192} \sin 6x + C$$

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    Can I red it.?. – 2017-02-10
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    Sorry, you can rollback to erase red type. – 2017-02-10
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    tag lines with `\tag{#}` at the end. this make it simple reference. – 2017-02-10
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    $\cos4x\cos2x\neq\cos6x$ – 2017-02-10
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    @MyGlasses I saw that comment too late but thanks, the red made me see what I was missing, a $\frac 1{32}\sin 2x$ will pop out and it is just what I needed! I'm sorry what do you mean by tag lines? – 2017-02-10
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    Sorry about my edit. I tagged last line. – 2017-02-10
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    Did you mean $\frac{3}{64}\sin{4x}$ for that third term? – 2017-02-10
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    @Triatticus Indeed I meant $\frac 3{64} \sin 4x$ thank you for pointing that out I edited it. – 2017-02-10
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    bashing integration by parts man rip – 2017-02-10

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When simplifying the split integral, $\cos^2 2x$ became $\frac{\cos{4x}}{2}$. I think you missed out something here.