Question

Can you help me with this integral by parts and explain me x^3cos^4(x/8) dx

Answer 1

Are you sure that's the correct integral? Because this one looks really nasty

Answer 2

There's a variety of ways to approach this. I'd suggest using the half-angle formula to reduce the power of the cosine.
\[\begin{align*}\cos^4\frac{x}{8}&=\left(\frac{1+\cos\dfrac{2x}{8}}{2}\right)^2\\\\
&=\frac{1}{4}+\frac{1}{2}\cos\frac{x}{4}+\frac{1}{4}\cos^2\frac{x}{4}\\\\
&=\frac{1}{4}+\frac{1}{2}\cos\frac{x}{4}+\frac{1}{4}\left(\frac{1+\cos\dfrac{2x}{4}}{2}\right)\\\\
&=\frac{3}{8}+\frac{1}{2}\cos\frac{x}{4}+\frac{1}{8}\cos\frac{x}{2}
\end{align*}\]
Integral:
\[\begin{align*}
\int x^3\cos^4\frac{x}{8}~dx&=\frac{3}{8}\int x^3~dx+\frac{1}{2}\int x^3\cos\frac{x}{4}~dx+\frac{1}{8}\int x^3\cos\frac{x}{2}~dx\\\\
&=\frac{3}{32}x^4+\frac{1}{2}\int x^3\cos\frac{x}{4}~dx+\frac{1}{8}\int x^3\cos\frac{x}{2}~dx
\end{align*}\]
For the cosine integrals, when you integrate by parts be sure to use
\[\begin{matrix}
u=x^3&&&dv=\cos\frac{x}{n}\\\\
du=3x^2~dx&&&v=n\sin\frac{x}{n}
\end{matrix}\]
where \(u\) will be the power of \(x\) and \(v\) will be the trig factor.