Find $d$ when $(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$ There exist constants $a$, $b$, $c$, and $d$ such that
$(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$
for all angles $x$. Find $d$.

This is the Fourier series for $f(x) = \sin^7 x$ on the domain $[0,2 \pi]$.
You have $\int_0^{2 \pi} \sin^7 x \sin x dx = d\int_0^{2 \pi} \sin x \sin x dx$.
Here is the answer:

 This gives ${35 \over 64 } \pi = d \pi$.

To compute the integral,
\begin{eqnarray}
\int_0^{2 \pi} \sin^8 x &=& { 1\over (2i)^8} \int_0^{2 \pi} (e^{it}-e^{-it})^8 dt \\
&=& { 1\over (2i)^8} \sum_{k=0}^8 \binom{8}{k} \int_0^{2 \pi} e^{it(8-2k)} dt \\
&=& { 1\over (2i)^8} \binom{8}{4} \int_0^{2 \pi} 1 dt \\
&=& {35 \over 64} \pi
\end{eqnarray}
