Question

Integrate $\int_{}^{}{\sin^{8}xdx}$

• A. $\frac{1}{2^{7}}\left\lbrack \frac{\sin 8x}{8} - 8\frac{\sin 6x}{6} + 28\frac{\sin 4x}{4} - 56\frac{\sin 2x}{2} + 35x + c \right\rbrack$
• B. $\frac{1}{2^{7}}\left\lbrack - \frac{\sin 8x}{8} + 8\frac{\sin 6x}{6} + 28\frac{\sin 4x}{4} + 56\frac{\sin 2x}{2} - 35x + c \right\rbrack$
• C. $\frac{\sin 8x}{8} - 8\frac{\sin 6x}{6} + 28\frac{\sin 4x}{4} - 56\frac{\sin 2x}{2} + 35x + c$
• D. None of these

Answer 1

Answer: (A)

$\frac{1}{2^{7}}\left\lbrack \frac{\sin 8x}{8} - 8\frac{\sin 6x}{6} + 28\frac{\sin 4x}{4} - 56\frac{\sin 2x}{2} + 35x + c \right\rbrack$

Answer 2

Let$\cos x + i \sin x = y$; then

$2 \cos x = y + \frac { 1 } { y } , 2 \cos n x = y ^ { n } + \frac { 1 } { y ^ { n } }$ $\mathbf{\Rightarrow}\mathbf{2i}\mathbf{\sin}\mathbf{x}\mathbf{= y}\mathbf{-}\frac{\mathbf{1}}{\mathbf{y}}\mathbf{,2i}\mathbf{\sin}\mathbf{n}\mathbf{x =}\mathbf{y}^{\mathbf{n}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{y}^{\mathbf{n}}}$ (Remember as the

standard results)

Thus $2 ^ { 8 } i ^ { 8 } \sin ^ { 8 } x = \left( y - \frac { 1 } { y } \right) ^ { 8 }$ $= \left( y^{8} + \frac{1}{y^{8}} \right) - 8\left( y^{6} + \frac{1}{y^{6}} \right) + 28\left( y^{4} + \frac{1}{y^{4}} \right) - 56\left( y^{2} + \frac{1}{y^{2}} \right) + 70$ $= 2 \cos 8 x - 16 \cos 6 x + 56 \cos 4 x - 112 \cos 2 x + 70$Thus $\sin ^ { 8 } x = \frac { 1 } { 2 ^ { 7 } } ( \cos 8 x - 8 \cos 6 x + 28 \cos 4 x - 56 \cos 3 x + 35 )$, and $\int \sin ^ { 8 } x d x = \frac { 1 } { 2 ^ { 7 } } \left[ \frac { \sin 8 x } { 8 } - 8 \frac { \sin 6 x } { 6 } + 28 \frac { \sin 4 x } { 4 } - 56 \frac { \sin 2 x } { 2 } + 35 x \right] + c$