Let I = \(\int_0^{2\pi}sin^7\frac{x}{4}{d}x
\)
= \(\int_0^{2\pi}(sin^2\frac{x}{4})^3sin\frac{x}{4}{d}x
\)
= \(\int_0^{2\pi}(1-cos^2\frac{x}{4})^3sin\frac{x}{4}{d}x
\)
Let cos\(\frac{x}{4}\) = t
⇒ \(-\frac{1}{4}\) sin\(\frac{x}{4}\) dx = dt
⇒ sin\(\frac{x}{4}\)dx = -4dt
Limits changes to t = cos 0 = 1 to t = cos\(\frac{2\pi}{4}\)
= cos\(\frac{\pi}{2}\) = 0
∴ I = \(\int_1^0 \) (1-t2) x -4dt
= 4\(\int_0^1 \)(1-t2)3 dt
= 4\(\int_0^1 \)(1-3t2+3t4-t6)dt
= 4 [t - 3\(\frac{t^3}{3}\)+ 3\(\frac{t^5}{5}\) - \(\frac{t^7}{7}\)\(]^1_0\)
= 4(1-1+\(\frac{3}{5}\)- \(\frac{1}{7}\))
= 4 x \(\frac{21-5}{35}\)
= \(\frac{64}{35}\).
