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(vi) \( \int_{0}^{2 \pi} \sin ^{7} \frac{x}{4} d x \)

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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 \(\frac{21-5}{35}\) 

\(\frac{64}{35}\).

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