Help with $\int \cos^6{(x)} \,dx$ Problem:
\begin{eqnarray*}
\int \cos^6{(x)} dx \\
\end{eqnarray*}
Answer:
\begin{eqnarray*}
\int \cos^4{(x)} \,\, dx &=& \int { \cos^2{(x)}(\cos^2{(x)})  } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=& \int { \frac{(1+\cos(2x))^2}{4}  } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=&
    \int { \frac{\cos^2(2x)^2 + 2\cos(2x)+1}{4}  } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=&
    \int { \frac{(\frac{1+\cos(4x)}{2} + 2\cos(2x)+1}{4}  } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=&
    \int { \frac{1+\cos(4x) + 4\cos(2x)+2}{8}  } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=&
    \int { \frac{\cos(4x) + 4\cos(2x)+3}{8}  } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=& \frac{\sin(4x)+ 8 \sin(2x)+12x}{32} \\
\text{Let }I_6 &=& \int \cos^6{(x)} \,\, dx \\
\end{eqnarray*}
To perform this integration, I use integration by parts with
$u = \cos^5(x)$ and $dv = \cos(x) dx$.
\begin{eqnarray*}
I_6 &=& \sin(x)\cos^5(x) - \int \sin(x) 5\cos^4(x)(-\sin(x)) \,\, dx \\
I_6 &=& \sin(x)\cos^5(x) + \int 5\cos^4(x)(\sin(x))^2 \,\, dx \\
I_6 &=& \sin(x)\cos^5(x) + \int 5\cos^4(x)(1 - \cos(x))^2 \,\, dx \\
I_6 &=& \sin(x)\cos^5(x) + \int 5\cos^4(x) \,\, dx  - 5I_6 \\
6I_6 &=& \sin(x)\cos^5(x) + \int 5\cos^4(x) \,\, dx \\
6I_6 &=& \sin(x)\cos^5(x) + \frac{5\sin(4x)+ 40 \sin(2x)+60x}{32}  + C_1 \\
6I_6 &=& \frac{32\sin(x)\cos^5(x) + 5\sin(4x)+ 40 \sin(2x)+60x}{32} + C_1 \\
I_6 &=& \frac{32\sin(x)\cos^5(x) + 5\sin(4x)+ 40 \sin(2x)+60x}{192} + C  \\
\end{eqnarray*}
I believe that the above result is wrong. Using an online integral
calculator, I get:
\begin{eqnarray*}
I_6 &=& \frac{\sin(6x) + 9\sin(4x) + 45 \sin(2x) + 60x}{192} + C \\
\end{eqnarray*}
I am hoping that somebody can tell me where I went wrong.
Bob

In your solution, when substituting the already known expression for $I_4$ (in the third line from the bottom), you forgot to multiply it by $5$. That's the only error there. Put it back in there, and you'll have a correct answer. Your answer would still look different from the output of that online integrator, but the two are in fact equivalent via trigonometric identities.
On a side note, this integral can also be found without integration by parts, but by using the same approach that worked for $I_4$ if you write $\cos^6(x)=(\cos^2(x))^3$.
