Computing the trigonometric integral $\int \cos^6x \, dx$

Question

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, $$\frac 18 \int 1+3\cos 2x \, dx\;+\; \frac 3{8}\int \cos^2 2x\,dx \;+\; \frac 1{8} \int \cos^3 2x\, dx$$ $$= \, \frac 18 \int 1+3\cos 2x \, dx\;+\; \frac 3{16}\int 1+\cos 4x\,dx \;+\; \frac 1{8}\int \cos^2 2x \cos 2x \, dx$$ $$= \, \frac 18 \int 1+3\cos 2x \, dx\;+\; \frac 3{16}\int 1+\cos 4x\,dx \;+\; \frac 1{16}\int \left(1+\cos 4x\right) \cos 2x \, dx$$

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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, $$= \, \frac 5{16}x + \frac {15}{64}\sin 2x + \frac 3{64} \sin 4x + \frac 1{192} \sin 6x + C$$

Answer 1

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

Answer 2

I find this expansion pretty useful for evaluating integrals related to powers of sin/cos:

$\displaystyle \begin{array}{{>{\displaystyle}l}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ To\ expand\ integer\ powers\ ( n) \ of\ cos^{n}(x) \ or\ sin^{n}( x ) :\\ \\ For\ odd\ n:\ cos^{n}( x ) =\left(\frac{e^{i x } +e^{-i x }}{2^{n}}\right)^{n} =\sum _{k=0}^{\left\lfloor \frac{n}{2}\right\rfloor }\binom{n}{k} \cdotp \frac{cos( x ( n-2k))}{2^{n-1}}\\ For\ even\ n:\ cos^{n}( x) =\left(\frac{e^{i x } +e^{-i x }}{2^{n}}\right)^{n} =\sum _{k=0}^{\left\lfloor \frac{n}{2}\right\rfloor }\binom{n}{k} \cdotp \frac{cos( x ( n-2k))}{2^{n-1}} -\frac{\left( 2\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right)\right) !}{2^{n} \cdot \left(\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right) !\right)^{2}}\\ \\ For\ odd\ n:\ \ sin^{n}( x ) =\left(\frac{e^{i x } -e^{-i x }}{( 2i)^{n}}\right)^{n} =\sum _{k=0}^{\left\lfloor \frac{n}{2}\right\rfloor }\binom{n}{k} \cdotp \frac{( -1)^{\left\lfloor \frac{n}{2}\right\rfloor -\ k} \cdot sin( x ( n-2k))}{2^{n-1}}\\ For\ even\ n:\ sin^{n}( x ) =\left(\frac{e^{ix } -e^{-ix }}{( 2i)^{n}}\right)^{n} =\sum _{k=0}^{\left\lfloor \frac{n}{2}\right\rfloor }\binom{n}{k} \cdotp \frac{( -1)^{\left\lfloor \frac{n}{2}\right\rfloor -\ k} \cdot cos(x ( n-2k))}{2^{n-1}} -\frac{\left( 2\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right)\right) !}{2^{n} \cdot \left(\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right) !\right)^{2}}\\ \\ \lfloor n\rfloor \ is\ the\ \ Floor\ function:\ \binom{n}{k} \ is\ the\ Binomial\ Coefficient\ funtion:\binom{n}{k} =\frac{n!}{k!\cdot ( n-k) !} \end{array}$